Eamcet Mathematics - Logarithms and Their Properties Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
If log23=a, log35=b and log7c=c, then the logarithm of the number 63 to base 140 is
- (a)
\(\frac { 1+2ac }{ 2c+abc+1 } \)
- (b)
\(\frac { 1-2ac }{ 2c-abc-1 } \)
- (c)
\(\frac { 1-2ac }{ 2c+abc+1 } \)
- (d)
\(\frac { 1+2ac }{ 2c-abc-1 } \)
\(\because \quad a=\frac { \log { 3 } }{ \log { 2 } } ,b=\frac { \log { 5 } }{ \log { 3 } } \quad and\quad c=\frac { \log { 2 } }{ \log { 7 } } \)
\(\therefore \quad abc=\frac { \log { 5 } }{ \log { 7 } } ,2c=\frac { 2\log { 2 } }{ \log { 7 } } \quad and\quad ac=\frac { \log { 3 } }{ \log { 7 } } \quad \quad ...(i)\)
\(\therefore \quad \log _{ 140 }{ 63 } =\frac { \log { 63 } }{ \log { 140 } } =\frac { 2\log { 3 } +\log { 7 } }{ \log { 7 } +2\log { 2 } +\log { 5 } } \)
\(=\frac { 2\left( \frac { \log { 3 } }{ \log { 7 } } \right) +1 }{ 1+2\left( \frac { \log { 2 } }{ \log { 7 } } \right) +\frac { \log { 5 } }{ \log { 7 } } } \)
\(=\frac { 1+2ac }{ 2c+abc+1 } \) [from Eq. (i)]
Ifxn>xn-1>....x2>x1>1, then the value of logx1logx2logx3....logxn\(\log _{ xn }{ { x }_{ n }^{ { x }_{ n-1 }^{ { . }^{ { .x1 }^{ } } } } } \) is
- (a)
0
- (b)
1
- (c)
2
- (d)
undefined
\(\because\) logx1logx2logx3....logxn-1logxn\(\log _{ xn }{ { x }_{ n }^{ { x }_{ n-1 }^{ { . }^{ { .x1 }^{ } } } } } \)
=logx1logx2logx3....logxn-1\((\log _{ xn }{ { x }_{ n }^{ { x }_{ n-2 }^{ { . }^{ { .x1 }^{ } } } } } .\log _{ { x }_{ n } }{ { x }_{ n } } )\)
=logx1logx2logx3....logxn-1\(\log _{ xn }{ { x }_{ n }^{ { x }_{ n-2 }^{ { . }^{ { .x1 }^{ } } } } } \)
.........
=logx1x1=1.
The number of real values of the parameter \(\lambda\)for which (log16x)2-log16x+log16\(\lambda\)=0 with real coefficients will have exactly one solution is
- (a)
1
- (b)
2
- (c)
3
- (d)
4
\(\because \quad \lambda >0\)
\(\therefore \quad \log _{ 16 }{ x } =\frac { 1\pm \sqrt { (1-4\log _{ 16 }{ \lambda } ) } }{ 2 } \)
\(\therefore\) The given equation will have exactly will have one solution, if
1-4log16\(\lambda\)=0 or log16\(\lambda\)=\(\frac {1}{4}\)=4-1
\(\therefore \quad { (16) }^{ { 4 }^{ -1 } }={ ({ 2 }^{ 4 }) }^{ 1/4 }=2,-2,2i,-2i,\quad where\quad i=\sqrt { -1 } \)
But \(\lambda\)is real and positive
\(\therefore\) \(\lambda\)=2
Number of real values=1.
If \(y={ 2 }^{ \frac { 1 }{ \log _{ x }{ 4 } } }\) , then x is equal to
- (a)
\(\sqrt { y } \)
- (b)
y
- (c)
y2
- (d)
y3
\(\because \quad y={ 2 }^{ \frac { 1 }{ \log _{ x }{ 4 } } }\)
\(\Rightarrow \quad \log _{ 2 }{ y } =\frac { 1 }{ \log _{ x }{ 4 } } \) (\(\because\) x>0,x\(\ne\)1)
or log2y=log4x=\(\frac { 1 }{ 2 } \log _{ 2 }{ x } \)
or 2log2y=log2x
or log2y2=log2x
\(\therefore\) x=y2
if log2x+log2y\(\ge\)6, then the least value of x+y is
- (a)
4
- (b)
8
- (c)
16
- (d)
32
\(\therefore\) log2x+log2y\(\ge\)6
\(\Rightarrow\) log2(xy)\(\ge\)6
\(\therefore\) xy\(\ge\)26
or \(\sqrt {xy}\)\(\ge\)23
\(\because \quad \frac { x+y }{ 2 } \ge \sqrt { xy } \quad or\quad x+y\ge 2\sqrt { xy } \ge 16\)
\(\because \quad x+y\ge 16\)
A rational number which is 50 times its own logarithm to the base 10 is
- (a)
1
- (b)
10
- (c)
100
- (d)
1000
Let x be the rational number, then according to question,
x=50 x log10x
By trial x=100.
If log102,log10(2x+1), log10(2x+3) are in AP, then
- (a)
x=0
- (b)
x=1
- (c)
x=log102
- (d)
\(x=\frac { 1 }{ 2 } \log _{ 2 }{ 5 } \)
\(\because\) log102,log10(2x+1), log10(2x+3) are in AP.
\(\therefore\) 2,2x+1,2x+3 are in GP.
Then, (2x+1)=2(2x+3)
or 22x=5
or 2x=log25
or \(x=\frac { 1 }{ 2 } \log _{ 2 }{ 5 } \)
\(7\log { \left( \frac { 16 }{ 15 } \right) } +5\log { \left( \frac { 25 }{ 24 } \right) } +3\log { \left( \frac { 81 }{ 80 } \right) } \) is equal to
- (a)
0
- (b)
1
- (c)
log 2
- (d)
log 3
\(7\log { \left( \frac { { 2 }^{ 4 } }{ 5\times 3 } \right) } +5\log { \left( \frac { { 5 }^{ 2 } }{ { 2 }^{ 3 }\times 3 } \right) } +3\log { \left( \frac { { 3 }^{ 4 } }{ { 2 }^{ 4 }\times 5 } \right) } \)
=7{4log2-log5-log3}+5{2log5-3log2-log3}+3{4log3-4log2-log5}
=log2
If log103=0.477, the number of digit in 340 is
- (a)
18
- (b)
19
- (c)
20
- (d)
21
Let P=340
\(\therefore\) log10P=40log103=40 x 0.477
=19.08
\(\therefore\) Number of digits in 340=19+1=20
If x=log35, y=log1725, which one of the following is correct?
- (a)
x
- (b)
x=y
- (c)
x>y
- (d)
None of these
\(\frac { 1 }{ y } =\log _{ 25 }{ 17 } =\frac { 1 }{ 2 } \log _{ 5 }{ 17 } \)
and \(\frac { 1 }{ x } =\log _{ 5 }{ 3 } =\frac { 1 }{ 2 } \log _{ 5 }{ 9 } \)
\(\therefore \quad \frac { 1 }{ y } >\frac { 1 }{ x } \Rightarrow x>y\)
The number log27 is
- (a)
an integer
- (b)
a rational number
- (c)
an irrational number
- (d)
a prime number
Suppose log27 is a rational number, then
log27=\(\left( \frac { p }{ q } \right) \)
\(\Rightarrow\) 7=2p/q\(\Rightarrow\)7q=2p
odd=even (impossible)
\(\therefore\) log27 is an irrational number.
The domain of the function \(\sqrt { \left( \log _{ 0.5 }{ x } \right) } \)is
- (a)
(1,\(\infty\))
- (b)
(0,\(\infty\))
- (c)
(0,1]
- (d)
(0.5,1)
log0.5x\(\ge\)0 and x>0
Then, x\(\le\)1 and x>0
Then, x∈(0,1]
log10tan 10+log10tan 20+....+log10tan890 is equal to
- (a)
0
- (b)
1
- (c)
27
- (d)
81
log10{tan 10 tan 20 tan 30...tan 450....tan 870 tan 880 tan 890}
=log10{tan 10 tan 20 tan 30...tan 450....cot 30 cot 20 cot 10}
=log101=0
If log1227=a, then log616 is equal to
- (a)
\(2\left( \frac { 3-a }{ 3+a } \right) \)
- (b)
\(3\left( \frac { 3-a }{ 3+a } \right) \)
- (c)
\(4\left( \frac { 3-a }{ 3+a } \right) \)
- (d)
\(5\left( \frac { 3-a }{ 3+a } \right) \)
\(\because\) a=log1227=log12(3)3=3log123
\(=\frac { 3 }{ \log _{ 3 }{ 12 } } \)
\(=\frac { 3 }{ 1+\log _{ 3 }{ 4 } } \)
\(=\frac { 3 }{ 1+2\log _{ 3 }{ 2 } } \)
\(\therefore \quad \log _{ 3 }{ 2 } =\frac { 3-a }{ 2a } \)
Then, log616=log624=4log62
\(=\frac { 4 }{ \log _{ 3 }{ 6 } } \)
\(=\frac { 4 }{ 1+\log _{ 3 }{ 3 } } \)
\(=\frac { 4 }{ 1+\frac { 2a }{ 3-a } } \)
\(=4\left( \frac { 3-a }{ 3+a } \right) \)
log7log7\(\sqrt { 7\sqrt { (7\sqrt { 7 } ) } } \) is equal to
- (a)
3log27
- (b)
3log22
- (c)
1-3log22
- (d)
1-3log27
\(\log _{ 7 }{ \log _{ 7 }{ { 7 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } } } } =\log _{ 7 }{ \left( \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 8 } \right) } =\log _{ 7 }{ \left( \frac { 7 }{ 8 } \right) } \)
=1-log78
=1-3log72
If\(\frac { \log { x } }{ b-c } =\frac { \log { y } }{ c-a } =\frac { \log { z } }{ a-b } \) then xaybzc is equal to
- (a)
xyz
- (b)
abc
- (c)
0
- (d)
1
\(\because \quad \frac { \log { x } }{ b-c } =\frac { \log { y } }{ c-a } =\frac { \log { z } }{ a-b } \)
\(=\frac { a\log { x } +b\log { y } +c\log { z } }{ 0 } \)
\(\Rightarrow\) log(x a y b z c)=0 \(\Rightarrow\)x a y b z c=1
\(\frac { 1 }{ 1+\log _{ a }{ bc } } +\frac { 1 }{ 1+\log _{ b }{ ca } } +\frac { 1 }{ 1+\log _{ x }{ ab } } \) is equal to
- (a)
0
- (b)
1
- (c)
2
- (d)
3
\(\frac { 1 }{ 1+\log _{ a }{ bc } } +\frac { 1 }{ 1+\log _{ b }{ ca } } +\frac { 1 }{ 1+\log _{ x }{ ab } } \)
\(=\frac { \log { a } }{ \log { a } +\log { b } +\log { c } } +\frac { \log { b } }{ \log { a } +\log { b } +\log { c } } +\frac { \log { c } }{ \log { a } +\log { b } +\log { c } } \)
=1
If (4)log93+(9)log24=(10)logx83, then x is equal to
- (a)
2
- (b)
3
- (c)
10
- (d)
30
(4)log93+(9)log24=(10)logx83
\(\Rightarrow\) (4)1/2+92=(10)logx83
\(\Rightarrow\) (83)1=(83)logx83
\(\Rightarrow\) 1=logx10 \(\Rightarrow\)x=10
If x,y,z are in GP and ax=by=cz, then
- (a)
logba=logcb
- (b)
logcb=logac
- (c)
logac=logba
- (d)
logcb=2logac
ax=by=cz
\(\Rightarrow\) xloga=ylogb=zlogc
\(\therefore \quad \frac { y }{ x } =\frac { z }{ y } \Rightarrow \frac { \log { a } }{ \log { b } } =\frac { \log { b } }{ \log { c } } \)
\(\Rightarrow\) logba=logcb
If \(\frac { x(y+z-x) }{ \log { x } } +\frac { y(z+x-y) }{ \log { y } } +\frac { z(x+y-z) }{ \log { z } } \)then xyyx=zyyz is equal to
- (a)
zxxz
- (b)
xzyx
- (c)
xyyz
- (d)
xxyy
Let \(\frac { x(y+z-x) }{ \log { x } } +\frac { y(z+x-y) }{ \log { y } } +\frac { z(x+y-z) }{ \log { z } } \)
\(=\frac { z(x+y-z) }{ \log { z } } =\frac { 1 }{ \lambda } \)
\(\therefore\) logx=\(\lambda\)x(y+z-x), logy=\(\lambda\)y(z+x-y), logz=\(\lambda\)z(z(x+y-z)
\(\therefore\) y log x+x log y=z log x+x log z=y log z+z log y
\(\Rightarrow\) log(xyyx)=log(xzzx)=log(yzzy)
Hence, xyyx=xzzx=yzzy
If\(\log _{ \sqrt { 8 } }{ b } =3\frac { 1 }{ 3 } ,\) then b is equal to
- (a)
2
- (b)
8
- (c)
32
- (d)
64
\(\log _{ \sqrt { 8 } }{ b } =3\frac { 1 }{ 3 } \)
\(\Rightarrow \quad \frac { 2 }{ 3 } \log _{ 2 }{ b } =\frac { 10 }{ 3 } \Rightarrow \log _{ 2 }{ b } =5\)
\(\Rightarrow \quad b={ 2 }^{ 5 }=32\)
If log0.3(x-1)
- (a)
\((-\infty,1)\)
- (b)
(1,2)
- (c)
\((2,\infty)\)
- (d)
none of these
\(\because\) x-1>0 \(\Rightarrow\)x>1
and log0.3(x-1)>log(0.3)2(x-1)
\(\Rightarrow\) log0.3(x-1)>\(\frac {1}{2}\)log0.3(x-1)
\(\frac {1}{2}\)log0.3(x-1)>0 \(\Rightarrow\)log0.3(x-1)>0
\(\Rightarrow\) x-1<1, \(\therefore\)x<2
Then, x\(\epsilon\)(1,2)
The value of 3log45-5log43 is
- (a)
0
- (b)
1
- (c)
2
- (d)
none of these
3log45-5log43
=3log45-5log43=0
If \(\frac { 1 }{ \log _{ 3 }{ \pi } } +\frac { 1 }{ \log _{ 4 }{ \pi } } \)>x, then x be
- (a)
2
- (b)
3
- (c)
\(\pi\)
- (d)
none of these
\(\frac { 1 }{ \log _{ 3 }{ \pi } } +\frac { 1 }{ \log _{ 4 }{ \pi } } \)>x
\(\Rightarrow \quad \log _{ \pi }{ 3 } +\log _{ \pi }{ 4 } >x\)
\(\Rightarrow \quad \log _{ \pi }{ 12 } >x\quad \Rightarrow 12>{ \pi }^{ x }\)
\(\therefore\) x=2
If \(In\left( \frac { a+b }{ 3 } \right) =\left( \frac { Ina+Inb }{ 2 } \right) ,\)then \(\frac { a }{ b } +\frac { b }{ a } \) is equal to
- (a)
1
- (b)
3
- (c)
5
- (d)
7
\(In\left( \frac { a+b }{ 3 } \right) =\frac { In\quad ab }{ 2 } =In\quad \sqrt { ab } \)
\(\Rightarrow \quad \frac { a+b }{ 3 } =\sqrt { ab } \)
\(\Rightarrow \quad { a }^{ 2 }+2ab+{ b }^{ 2 }=9ab\)
\(\Rightarrow \quad \frac { a }{ b } +2+\frac { b }{ a } =9\)
\(\therefore \quad \frac { a }{ b } +\frac { b }{ a } =7\)
If log3{5+4log3(x-1)}=2, then x is equal to
- (a)
2
- (b)
4
- (c)
8
- (d)
log216
x-1>0 \(\Rightarrow\)x>1 ....(i)
and 5+4log3(x-1)>0
\(\Rightarrow\)4log3(x-1)>-5
\(\Rightarrow\) log3(x-1)>\(-\frac { 5 }{ 4 } \)
\(\Rightarrow\) x-1>3-5/4
\(\Rightarrow\) x-1>1+3-5/4 ....(ii)
From Eqs. (i) and (ii), x>1+3-5/4
\(\therefore\) 5+4log3(x-1)=9
\(\Rightarrow\) 4log3(x-1)=4
\(\Rightarrow\) log3(x-1)=1
\(\Rightarrow\) x-1=3
\(\therefore\) x=4
If 2xlog43+3log4x=27, then x is equal to
- (a)
2
- (b)
4
- (c)
8
- (d)
16
2xlog43+3log4x=27
\(\Rightarrow\) 2.3log4x+3log4x=27
\(\Rightarrow\) 3log4x=9=32
log4x=2
\(\therefore\) x=42=16
The solution set of the equation logx2log2x2=log4x2 is
- (a)
\(\left\{ { 2 }^{ -\sqrt { 2 } },{ 2 }^{ \sqrt { 2 } } \right\} \)
- (b)
{1/2,2}
- (c)
{1/4,22}
- (d)
none of these
\(\because\) logx2log2x2=log4x2
\(\because\) x>0,2x>0 and 4x>0 and
x\(\ne\)1, 2x\(\ne\)1, 4x\(\ne\)1
\(\Rightarrow\) x>0 and \(x\neq 1,\frac { 1 }{ 2 } ,\frac { 1 }{ 4 } \)
Then, \(\frac { 1 }{ \log _{ 2 }{ x } } .\frac { 1 }{ \log _{ 2 }{ 2x } } =\frac { 1 }{ \log _{ 2 }{ 4x } } \)
\(\Rightarrow \quad \log _{ 2 }{ x } .\log _{ 2 }{ 2x } =\log _{ 2 }{ 4x } \)
\(\Rightarrow \quad \log _{ 2 }{ x } .(1+\log _{ 2 }{ x } )=(2+\log _{ 2 }{ x } )\)
\(\Rightarrow \quad { \left( \log _{ 2 }{ x } \right) }^{ 2 }=2\)
\(\Rightarrow \quad \log _{ 2 }{ x } =\pm \sqrt { 2 } \)
\(\therefore \quad x={ 2 }^{ \pm \sqrt { 2 } }\)
\(\therefore \quad x=\left\{ { 2 }^{ -\sqrt { 2 } },{ 2 }^{ \sqrt { 2 } } \right\} \)
The solution of the equation \(\log _{ 7 }{ \log _{ 5 }{ \left( \sqrt { x+5 } +\sqrt { x } \right) } } =0\)is
- (a)
1
- (b)
3
- (c)
4
- (d)
5
It is clear that x\(\ge\)0 and x\(\ge\)-5
ie, x\(\ge\)0
\(\because \quad \log _{ 7 }{ \log _{ 5 }{ \left( \sqrt { x+5 } +\sqrt { x } \right) } } =0\)
\(\Rightarrow \quad \log _{ 5 }{ \left( \sqrt { x+5 } +\sqrt { x } \right) } =1\)
\(\Rightarrow \quad \sqrt { x+5 } +\sqrt { x } =5\)
\(\Rightarrow \quad \sqrt { (x+5) } =5-\sqrt { x } \)
or \((x+5)={ \left( 5-\sqrt { x } \right) }^{ 2 }\)
\(\Rightarrow \quad x+5=25+x-10\sqrt { x } \)
or \(10\sqrt { x } =20\)
\(\Rightarrow \quad \sqrt { x } =2\)
\(\therefore \quad x=4\)
It is given that x=9 is a solution of the equation In(x2-15a2)-(In(a-2)=\(In\left( \frac { 8ax }{ a-2 } \right) ,\) then
- (a)
a=3
- (b)
a=\(\frac {3}{5}\)
- (c)
other solution is x=3/5
- (d)
other solution is x=15
\(\because \quad a-2>\quad and\quad \frac { 8ax }{ a-2 } >0\)
Then, a>2 and x>0 ....(i)
Then, the given equation can be written as
In(x2-15a2)\(=In\left\{ \frac { 8ax }{ (a-2) } .(a-2) \right\} \)
or In(x2-15a2)=In (8ax)
or x2+15a2=In (8ax)
or x2+15a2=8ax
or x2-8ax+15a2=0
or (x-5a)(x-3a)=0
\(\therefore\) x=3a, 5a
But x=9, then 9=3a, 5a
\(\therefore\) a=3, 9/5
or a=3 (\(\because\) a>2)
Then, x=9, 15
The value of {logba.logcb.logdc.logad} is
- (a)
0
- (b)
log abcd
- (c)
log 1
- (d)
1
logba.logcb.logdc.logad
=logca.logdc.logad
=logda.logad
=1
If \(\frac { \log _{ 2 }{ x } }{ 4 } =\frac { \log _{ 2 }{ y } }{ 6 } =\frac { \log _{ 2 }{ z } }{ 3k } \)and x3y2z=1, then k is equal
- (a)
-8
- (b)
-4
- (c)
0
- (d)
\(\log _{ 2 }{ \left( \frac { 1 }{ 256 } \right) } \)
\(\frac { \log _{ 2 }{ x } }{ 4 } =\frac { \log _{ 2 }{ y } }{ 6 } =\frac { \log _{ 2 }{ z } }{ 3k } \)
\(=\frac { 3\log _{ 2 }{ x } +2\log _{ 2 }{ y } +\log _{ 2 }{ z } }{ 12+12+3k } \)
\(=\frac { \log _{ 2 }{ \left( { x }^{ 2 }{ y }^{ 2 }z \right) } }{ 24+3k } \)
\(=\frac { 0 }{ 24+3k } \)
\(\therefore \quad 24+3k=0,\quad then\quad k=-8\)
If \(\frac { \log { a } }{ (b-c) } =\frac { \log { b } }{ (c-a) } =\frac { \log { c } }{ (a-b) } ,\) then ab+c.bc+a.ca+b is equal to
- (a)
0
- (b)
1
- (c)
a+b+c
- (d)
logba.logcb.logac
\(\frac { (b+c)\log { a } +(c+a)\log { b } +(a+b)\log { c } }{ (b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b) } \)
\(=\frac { { \log { a } }^{ b+c }+{ \log { b } }^{ c+a }+{ \log { c } }^{ a+b } }{ 0 } \)
\(=\frac { \log { \left( a^{ b+c }.{ b }^{ c+a }.{ c }^{ a+b } \right) } }{ 0 } \)
\(\therefore \quad \log { \left( a^{ b+c }.{ b }^{ c+a }.{ c }^{ a+b } \right) } =0\)
Then, ab+c.bc+a.ca+b=1
The expression \({ 5 }^{ \log _{ 1/5 }{ \left( 1/2 \right) } }+\log _{ \sqrt { 2 } }{ \left( \frac { 4 }{ \sqrt { 7 } +\sqrt { 3 } } \right) } +\log _{ 1/2 }{ \left( \frac { 1 }{ 10+2\sqrt { 21 } } \right) } \) simplifies to
- (a)
6
- (b)
4
- (c)
\(\sqrt { 6\sqrt { 6\sqrt { 6\sqrt { 6.....\infty } } } } \)
- (d)
\({ 3 }^{ \log _{ 1/3 }{ \left( \frac { 1 }{ 6 } \right) } }\)
\(\because \quad { 5 }^{ \log _{ 1/5 }{ \left( 1/2 \right) } }={ 5 }^{ \log _{ 5 }{ \left( 2 \right) } }=2\)
\(\log _{ \sqrt { 2 } }{ \left( \frac { 4 }{ \sqrt { 7 } +\sqrt { 3 } } \right) } =\log _{ \sqrt { 2 } }{ \left( \frac { 4\left( \sqrt { 7 } -\sqrt { 3 } \right) }{ \sqrt { 7 } +\sqrt { 3 } } \right) } =2\log _{ 2 }{ \left( \sqrt { 7 } -\sqrt { 3 } \right) } \)
\(=\log _{ 2 }{ { \left( \sqrt { 7 } -\sqrt { 3 } \right) }^{ 2 } } =\log _{ 2 }{ { \left( 10-2\sqrt { 21 } \right) } } \)
and \(\log _{ 1/2 }{ \left( \frac { 1 }{ 10+2\sqrt { 21 } } \right) } =\log _{ 2 }{ { \left( 10+2\sqrt { 21 } \right) } } \)
\(\therefore\) Given expression can be written as
=2+\(\log _{ 2 }{ { \left( 10-2\sqrt { 21 } \right) } } \)\(+\log _{ 2 }{ { \left( 10+2\sqrt { 21 } \right) } } \)
\(=2+\log _{ 2 }{ \left\{ \left( 10-2\sqrt { 21 } \right) \left( 10+2\sqrt { 21 } \right) \right\} } \)
=2+log2(16)=2+4=6
Also, \(\sqrt { 6\sqrt { 6\sqrt { 6\sqrt { 6..... } } } } ={ 6 }^{ \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\frac { 1 }{ 6 } +.... }={ 6 }^{ \frac { \frac { 1 }{ 2 } }{ 1-1/2 } }=6\)
and \({ 3 }^{ \log _{ 1/3 }{ (1/6) } }={ 3 }^{ \log _{ 3 }{ (6) } }=6\)
If \(\log _{ a }{ x } =\alpha ,\log _{ b }{ x } =\beta ,\log _{ c }{ x } =\gamma \quad and\quad \log _{ d }{ x } =\delta ,\quad x\neq 1\quad and\quad a,b,c,d\neq 0,>q,\quad then\quad \log _{ abcd }{ x } \)equals
- (a)
\(\le \frac { \alpha +\beta +\gamma +\delta }{ 16 } \)
- (b)
\(\ge \frac { \alpha +\beta +\gamma +\delta }{ 16 } \)
- (c)
\(\frac { 1 }{ { \alpha }^{ -1 }+{ \beta }^{ -1 }+{ \gamma }^{ -1 }+{ \delta }^{ -1 } } \)
- (d)
\(\frac { 1 }{ \alpha \beta \gamma \delta } \)
\(\because \quad \log _{ abcd }{ x } =\frac { 1 }{ \log _{ abcd }{ x } } \)
\(=\frac { 1 }{ \log _{ a }{ x } +\log _{ b }{ x } +\log _{ c }{ x } +\log _{ d }{ x } } \)
\(=\frac { 1 }{ { \alpha }^{ -1 }+{ \beta }^{ -1 }+{ \gamma }^{ -1 }+{ \delta }^{ -1 } } \)
Also, AM\(\ge\) GM
\(\Rightarrow \quad \frac { \alpha +\beta +\gamma +\delta }{ 4 } \ge \frac { 4 }{ \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma } +\frac { 1 }{ \delta } } \)
\(\frac { 1 }{ { \alpha }^{ -1 }+{ \beta }^{ -1 }+{ \gamma }^{ -1 }+{ \delta }^{ -1 } } \le \frac { \alpha +\beta +\gamma +\delta }{ 16 } \)
\(\log _{ p }{ \log _{ p }{ \underbrace { \sqrt [ p ]{ \sqrt [ p ]{ \sqrt [ p ]{ ....\sqrt [ p ]{ p } } } } }_{ n\quad times } , } } \) p>0 and p\(\ne\)1, is equal to
- (a)
n
- (b)
-n
- (c)
\(\frac {1}{n}\)
- (d)
\(\log _{ 1/p }{ ({ p }^{ n }) } \)
\(\underbrace { \sqrt [ p ]{ \sqrt [ p ]{ \sqrt [ p ]{ ....\sqrt [ p ]{ p } } } } }_{ n\quad times } ={ p }^{ 1/{ p }^{ n } }={ p }^{ { p }^{ -n } }\)
\(\log _{ p }{ \log _{ p }{ \underbrace { \sqrt [ p ]{ \sqrt [ p ]{ \sqrt [ p ]{ ....\sqrt [ p ]{ p } } } } }_{ n\quad times } } } \)
\(\log _{ p }{ \log _{ p }{ \underbrace { \sqrt [ p ]{ \sqrt [ p ]{ \sqrt [ p ]{ ....\sqrt [ p ]{ p } } } } }_{ n\quad times } } } =\log _{ p }{ \log _{ p }{ { p }^{ { p }^{ -n } } } } =\log _{ p }{ ({ p }^{ ^{ -n } }) } =-n\)
Sum of the roots of the equation x+1=2log2(2x+3)-2log4(1980-2-x) is
- (a)
log112
- (b)
log211
- (c)
log11(0.5)
- (d)
\(\log _{ 0.5 }{ \left( \frac { 1 }{ 11 } \right) } \)
\(\because\) x+1=2log2(2x+3)-2log22(1980-2-x)
\(\Rightarrow\) x+1=2log2(2x+3)-2log2(1980-2-x)
\(\Rightarrow \quad x+1=\log _{ 2 }{ \left\{ \frac { { ({ 2 }^{ x }+3) }^{ 2 } }{ 1980-{ 2 }^{ -x } } \right\} } \)
\(\Rightarrow \quad { 2 }^{ x }.2=\frac { { 2 }^{ 2x }+6.{ 2 }^{ x }+9 }{ 1980-{ 2 }^{ -x } } \)
\(\Rightarrow \quad 2\times 1980\times { 2 }^{ x }-2={ 2 }^{ 2x }+6.{ 2 }^{ x }+9\)
\(\Rightarrow \quad { 2 }^{ 2x }-3954.{ 2 }^{ x }+11={ 0 }\)
If roots are \(\alpha ,\beta ,\) then
\({ 2 }^{ \alpha }.{ 2 }^{ \beta }=\frac { 1 }{ 11 } \)
\(\Rightarrow \quad { 2 }^{ \alpha +\beta }=11\)
or \(\alpha +\beta =\log _{ 2 }{ 11 } \)
Also,
\(\log _{ 0.5 }{ \left( \frac { 1 }{ 11 } \right) } =\log _{ 1/2 }{ \left( \frac { 1 }{ 11 } \right) } =\log _{ 2 }{ 11 } \)
If ax=b, by=c, cz=a, x=logbak1,y=logcbk2,z=logack3, then k1k2k3 is equal to
- (a)
1
- (b)
abc
- (c)
(xyz)
- (d)
0
ax=b \(\Rightarrow\)logab
by=c \(\Rightarrow\)logbc
and cz=a \(\Rightarrow\)logca
\(\therefore\) xyz=1
and x=k1logba,y=k2logcb,z=k3logac
\(\therefore\) xyz=k1k2k3
or k1k2k3=xyz=1
If \(\frac { \ln { a } }{ b-c } =\frac { \ln { b } }{ c-a } =\frac { \ln { c } }{ a-b } ,\) a,b,c>0, then
- (a)
ab+c.bc+a.ca+b=1
- (b)
ab+c.bc+a.ca+b ≥3
- (c)
ab+c.bc+a.ca+b=3
- (d)
ab+c.bc+a.ca+b ≥ 3(3)1/3
Since, a>0, b>0, c>0
\(\Rightarrow \quad \frac { (b+c)\ln { a } }{ (b+c)(b-c) } =\frac { (c+a)\ln { b } }{ (c+a)(c-a) } =\frac { (a+b)\ln { c } }{ (a+b)(a-b) } \)
\(=\frac { (b+c)\ln { a } +(c+a)\ln { b } +(a+b)\ln { c } }{ 0 } \)
\(=\frac { \ln { { a }^{ b+c } } +\ln { { b }^{ c+a } } +\ln { { c }^{ a+b } } }{ 0 } \)
\(\Rightarrow \quad \ln { ({ a }^{ b+c }.{ b }^{ c+a }.{ c }^{ a+b }) } \)
Again, AM\(\ge\) GM
\(\Rightarrow \quad \frac { { a }^{ b+c }+{ b }^{ c+a }+{ c }^{ a+b } }{ 3 } \ge { ({ a }^{ b+c }.{ b }^{ c+a }.{ c }^{ a+b }) }^{ 1/3 }\)
\(={ (1) }^{ 1/3 }\)
=1
\(\therefore \quad { a }^{ b+c }+{ b }^{ c+a }+{ c }^{ a+b }\ge 3\)
If \(\frac { \ln { a } }{ y-z } =\frac { \ln { b } }{ z-x } =\frac { \ln { c } }{ x-y } ,\) then
- (a)
ax.by.cz=1
- (b)
ay2+yz+z2.bz2+zx+x2.cx2+xy+y2=1
- (c)
aa+b.bc+a.ca+b=1
- (d)
abc=1
\(\because \quad \frac { \ln { a } }{ y-z } =\frac { \ln { b } }{ z-x } =\frac { \ln { c } }{ x-y } \)
\(=\frac { \ln { a } +\ln { b } +\ln { c } }{ 0 } \)
\(=\frac { \ln { abc } }{ 0 } \)
or In(abc)=0 \(\Rightarrow\)abc=e0=1
\(\frac { \ln { a } }{ y-z } =\frac { \ln { b } }{ z-x } =\frac { \ln { c } }{ x-y } \)
\(=\frac { x\ln { a } +y\ln { b } +z\ln { c } }{ 0 } \)
\(=\frac { \ln { \left( { a }^{ x }.{ b }^{ y }.{ c }^{ z } \right) } }{ 0 } \)
or \(\ln { \left( { a }^{ x }.{ b }^{ y }.{ c }^{ z } \right) } =0\)
\(\Rightarrow \quad { a }^{ x }{ b }^{ y }{ c }^{ z }={ e }^{ 0 }=1\)
\(\frac { \ln { a } }{ y-z } =\frac { \ln { b } }{ z-x } =\frac { \ln { c } }{ x-y } \)
\(=\frac { (y+z)\ln { a } +(z+x)\ln { b } +(x+y)\ln { c } }{ 0 } \)
\(=\frac { \ln { \left\{ { a }^{ y+z }.{ b }^{ x+z }.{ c }^{ x+y } \right\} } }{ 0 } \)
\(or\quad \ln { \left\{ { a }^{ y+z }.{ b }^{ x+z }.{ c }^{ x+y } \right\} } =0\)
\(\Rightarrow \quad { a }^{ y+z }.{ b }^{ x+z }.{ c }^{ x+y }={ e }^{ 0 }=1\)
and \(\frac { \ln { a } }{ y-z } =\frac { \ln { b } }{ z-x } =\frac { \ln { c } }{ x-y } \)
\(=\frac { \left( { y }^{ 2 }+yz+{ z }^{ 2 } \right) \ln { a } +\left( { z }^{ 2 }+zx+{ x }^{ 2 } \right) \ln { b } +\left( { x }^{ 2 }+xy+{ y }^{ 2 } \right) \ln { c } }{ 0 } \)
\(=\frac { \ln { \left\{ { a }^{ { y }^{ 2 }+yz+{ z }^{ 2 } }.{ b }^{ { z }^{ 2 }+zx+{ x }^{ 2 } }.{ c }^{ { x }^{ 2 }+xy+{ y }^{ 2 } } \right\} } }{ 0 } \)
\(or\quad \ln { \left\{ { a }^{ { y }^{ 2 }+yz+{ z }^{ 2 } }.{ b }^{ { z }^{ 2 }+zx+{ x }^{ 2 } }.{ c }^{ { x }^{ 2 }+xy+{ y }^{ 2 } } \right\} } =0\)
\(\Rightarrow \quad { a }^{ { y }^{ 2 }+yz+{ z }^{ 2 } }.{ b }^{ { z }^{ 2 }+zx+{ x }^{ 2 } }.{ c }^{ { x }^{ 2 }+xy+{ y }^{ 2 } }={ e }^{ 0 }=1\)
The solution of the equation 3logax+3xloga3=2 is given by
- (a)
alog3a
- (b)
(2/a)log32
- (c)
a-log32
- (d)
2-log3a
\(\Rightarrow\) 3logax+3xloga3=2
\(\Rightarrow\) 3logax+3.3logax=2
\(\Rightarrow\) 4.3logax=2
\(\Rightarrow \quad { 3 }^{ \log _{ s }{ x } }=\left( \frac { 1 }{ 2 } \right) \)
\(\Rightarrow \log _{ 3 }{ x } =-\log _{ 3 }{ 2 } \Rightarrow x{ \log _{ 3 }{ x } }^{ -\log _{ 3 }{ ({ 2 }^{ -1 }) } }\)
\(={ ({ 2 }^{ -1 }) }^{ \log _{ 3 }{ a } }={ 2 }^{ -\log _{ 3 }{ a } }\)
An equation \(\begin{cases} f(x)>0 \\ { f }^{ 2m }(x)=g(x) \end{cases}\) is equivalent to the system 2mlogaf(x)=logag(x), a>0, a\(\ne\)1, m∈N. The number of solutions of 2loge2x=2loge(7x-2-2x2) is
- (a)
1
- (b)
2
- (c)
3
- (d)
infinite
This equation is equivalent to the system
\(\begin{cases} 2x>0 \\ { (2x })^{ 2 }=7x-2-{ 2x }^{ 2 }g(x) \end{cases}\)
\(\Rightarrow \quad \begin{cases} x>0 \\ (x-1/2)(x-2/3)=0 \end{cases}\)
\(\Rightarrow \quad \begin{cases} x=\frac { 1 }{ 2 } \\ x=\frac { 2 }{ 3 } \end{cases}\)
\(\therefore\) Number of solutions=2.
An equation \(\begin{cases} f(x)>0 \\ { f }^{ 2m }(x)=g(x) \end{cases}\) is equivalent to the system 2mlogaf(x)=logag(x), a>0, a\(\ne\)1, m∈N. The number of solutions of In 2x=2In (4x-15) is
- (a)
0
- (b)
1
- (c)
2
- (d)
infinite
This equation is equivalent to the system
\(\Rightarrow \quad \begin{cases} 4x-15>0 \\ 2x={ (4x-15) }^{ 2 } \end{cases}\)
\(\Rightarrow \quad \begin{cases} x>\frac { 15 }{ 4 } \\ 16x^{ 2 }-122x+225=0 \end{cases}\)
\(\therefore \quad x=\frac { 9 }{ 2 } \)
\(\therefore\) Number of solution=1.
An equation \(\begin{cases} f(x)>0 \\ { f }^{ 2m }(x)=g(x) \end{cases}\) is equivalent to the system 2mlogaf(x)=logag(x), a>0, a\(\ne\)1, m∈N. Solution set of the equation \(\log _{ ({ x }^{ 3 }+6) }{ ({ x }^{ 2 }-1) } =\log _{ ({ 2x }^{ 2 }+5x) }{ ({ x }^{ 2 }-1) } \) is
- (a)
{-2}
- (b)
{1}
- (c)
{3}
- (d)
{-2,1,3}
This equation is equivalent to the system
\(\begin{cases} { x }^{ 2 }-1>0 \\ 2{ x }^{ 2 }+5x>0,\quad \neq 1 \\ { x }^{ 3 }+6=2{ x }^{ 2 }+5x \end{cases}\)
\(\Rightarrow \quad \begin{cases} { x }<-1\quad and\quad x>1 \\ x<-\frac { 5 }{ 2 } \quad and\quad x>0 \\ x\neq \frac { -5\pm \sqrt { 33 } }{ 4 } \\ x=-2,1,3 \end{cases}\)
For final solution \(x<-\frac { 5 }{ 2 } \) and x>1
ie, only x=3.
An equation \(\begin{cases} f(x)>0 \\ { f }^{ 2m }(x)=g(x) \end{cases}\) is equivalent to the system 2mlogaf(x)=logag(x), a>0, a\(\ne\)1, m∈N. Solution set of the equation \(\log { (x-9) } +2\log { \sqrt { (2x-1) } } =2\) is
- (a)
\(\left\{ \phi \right\} \)
- (b)
{1}
- (c)
{2}
- (d)
{13}
x-9>0, 2x-1>0 and (x-9)(2x-1)=102=100
or x>9 and 2x2-19x-91=0
or x>9 and 2x2-26x+7x-91=0
or x=13.
Equations of the form (i) f(logax)=0, a>0, a\(\ne\)1 and (ii) g(logxA)=0, A>0, then Eq. (i) is equivalent to f(t)=0, where t=logax. If t1,t2,t3,....,tk are the roots of f(t)=0, then logax=t2,...,logax=tk and Eq. (ii) is equivalent to f(y)=0, where y=logxA. If y1,y2,y3,....,yk are the roots of f(y)=0, then logxA=y1, logxA=y2,....,logxA=yk. The number of solutions of the equation \(\frac { 1-2{ \left( \log { { x }^{ 2 } } \right) }^{ 2 } }{ \log { { x } } -2{ \left( \log { { x }^{ 2 } } \right) }^{ 2 } } =1\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
infinite
The given equation can rewrite in the form
\(\frac { 1-2{ \left( 2\log { { x }^{ 2 } } \right) }^{ 2 } }{ \log { { x } } -2{ \left( \log { { x }^{ 2 } } \right) }^{ 2 } } =1\)
\(\Rightarrow \quad \frac { 1-8{ \left( \log { { x } } \right) }^{ 2 } }{ \log { { x } } -2{ \left( \log { { x } } \right) }^{ 2 } } =1\)
Let log x=t
Then, \(\frac { { (1-8{ t }^{ 2 }) }^{ 2 } }{ t-2{ t }^{ 2 } } -1=0\)
\(\Rightarrow \quad \frac { 1-t-6{ t }^{ 2 } }{ t-2{ t }^{ 2 } } =0\)
\(\Rightarrow \quad \frac { \\ (1+2t)(1-3t) }{ t(1-2t) } =0\)
\(\Rightarrow \quad \begin{cases} t=-\frac { 1 }{ 2 } \\ t=\frac { 1 }{ 3 } \end{cases}\Rightarrow \begin{cases} log\quad x=-\frac { 1 }{ 2 } \\ log\quad x=\frac { 1 }{ 3 } \end{cases}\)
\(\Rightarrow \quad \begin{cases} x={ \left( 10 \right) }^{ -1/2 } \\ x={ (10) }^{ 1/3 } \end{cases}\)
\(\therefore\) Number of solution is 2.
Equations of the form (i) f(logax)=0, a>0, a\(\ne\)1 and (ii) g(logxA)=0, A>0, then Eq. (i) is equivalent to f(t)=0, where t=logax. If t1,t2,t3,....,tk are the roots of f(t)=0, then logax=t2,...,logax=tk and Eq. (ii) is equivalent to f(y)=0, where y=logxA. If y1,y2,y3,....,yk are the roots of f(y)=0, then logxA=y1, logxA=y2,....,logxA=yk. The number of solutions of the equation \(\log { _{ x }^{ 3 } } 10-6\log { _{ x }^{ 2 } } 10+11\log _{ x }{ 10 } -6=0\) is
- (a)
0
- (b)
1
- (c)
2
- (d)
3
Put logx10=t in the given equation, we get t3-6t2+11t-6=0
\(\Rightarrow\) (t-1)(t-2)(t-3)=0
Then, \(\begin{cases} t=1 \\ t=2, \\ t=3 \end{cases}if\quad follows\quad that\quad \begin{cases} \log _{ x }{ 10 } =1 \\ \log _{ x }{ 10 } =2 \\ \log _{ x }{ 10 } =3 \end{cases}\)
\(\Rightarrow \quad \begin{cases} x=10 \\ { x }^{ 2 }={ 10 } \\ { x }^{ 3 }=10 \end{cases}\quad \Rightarrow \quad \begin{cases} x=10 \\ x=\sqrt { 10 } \\ x=\sqrt [ 3 ]{ 10 } \end{cases}\quad \quad \quad (\because \quad x>0\quad and\quad x\neq 1)\)
\(\therefore\) Number of solution is 3.
Equations of the form (i) f(logax)=0, a>0, a\(\ne\)1 and (ii) g(logxA)=0, A>0, then Eq. (i) is equivalent to f(t)=0, where t=logax. If t1,t2,t3,....,tk are the roots of f(t)=0, then logax=t2,...,logax=tk and Eq. (ii) is equivalent to f(y)=0, where y=logxA. If y1,y2,y3,....,yk are the roots of f(y)=0, then logxA=y1, logxA=y2,....,logxA=yk. The solution set of \({ \left( \log _{ 5 }{ x } \right) }^{ 2 }+\log _{ 5 }{ x } +1=\frac { 7 }{ \log _{ 5 }{ x } -1 } \)contains
- (a)
(1,3)
- (b)
{1}
- (c)
{25}
- (d)
{1,25}
Put log5x=t in the given equation, we get
\({ t }^{ 2 }+t+1=\frac { 7 }{ t-1 } \)
\(\Rightarrow \quad { t }^{ 2 }+t+1-\frac { 7 }{ t-1 } =0\)
\(\Rightarrow \quad \frac { ({ t }^{ 3 }-1)-7 }{ t-1 } =0\)
\({ t }^{ 3 }=8,t\neq 1\)
\(\therefore \quad t=2\quad \Rightarrow \quad \log _{ 5 }{ x } =2\quad \Rightarrow x=5^{ 2 }=25\)
Equations of the form (i) f(logax)=0, a>0, a\(\ne\)1 and (ii) g(logxA)=0, A>0, then Eq. (i) is equivalent to f(t)=0, where t=logax. If t1,t2,t3,....,tk are the roots of f(t)=0, then logax=t2,...,logax=tk and Eq. (ii) is equivalent to f(y)=0, where y=logxA. If y1,y2,y3,....,yk are the roots of f(y)=0, then logxA=y1, logxA=y2,....,logxA=yk. The set of all x satisfying the equation \({ x }^{ \log _{ 3 }{ { x }^{ 2 } } +{ (\log _{ 3 }{ x } ) }^{ 2 } }=\frac { 1 }{ { x }^{ 2 } } \) is
- (a)
{1,9}
- (b)
\(\left\{ 9,\frac { 1 }{ 81 } \right\} \)
- (c)
\(\left\{ 1,\quad 4,\quad \frac { 1 }{ 81 } \right\} \)
- (d)
\(\left\{ 1,\quad 9,\quad \frac { 1 }{ 81 } \right\} \)
Taking log of both sides with base 3, we have {log3x2+(log3x)2-10}log3x=-2log3x
Put log3x=t, then (2t+t2-10+2)t=0
\(\Rightarrow \quad ({ t }^{ 2 }+2t-8)t=0\)
\(\Rightarrow\) (t+4)(t-2)t=0
\(\Rightarrow \quad \begin{cases} t=-4 \\ t=2 \\ t=0 \end{cases}\Rightarrow \quad \begin{cases} \log _{ 3 }{ x } =-4 \\ \log _{ 3 }{ x } =2 \\ \log _{ 3 }{ x } =0 \end{cases}\)
\(\Rightarrow \quad \begin{cases} x={ 3 }^{ -4 }=\frac { 1 }{ 81 } \\ x={ 3 }^{ 2 }=9 \\ x={ 3 }^{ 0 }=1 \end{cases}\)
\(\therefore\) Solution set is \(\left\{ 1,\quad 9,\quad \frac { 1 }{ 81 } \right\} \)
Equations of the form (i) f(logax)=0, a>0, a\(\ne\)1 and (ii) g(logxA)=0, A>0, then Eq. (i) is equivalent to f(t)=0, where t=logax. If t1,t2,t3,....,tk are the roots of f(t)=0, then logax=t2,...,logax=tk and Eq. (ii) is equivalent to f(y)=0, where y=logxA. If y1,y2,y3,....,yk are the roots of f(y)=0, then logxA=y1, logxA=y2,....,logxA=yk. If \(\frac { { (In\quad x) }^{ 2 }-3Inx+3 }{ Inx-1 } <1,\) then x belongs to
- (a)
(0,e)
- (b)
(1,e)
- (c)
(1,2e)
- (d)
(0,3e)
Put x=t, then
\(\frac { { t }^{ 2 }-3t+3 }{ t-1 } <1\)
\(\Rightarrow \quad \frac { { t }^{ 2 }-3t+3 }{ t-1 } -1<0\)
\(\Rightarrow \quad \frac { { t }^{ 2 }-4t+4 }{ t-1 } <0\)
\(\Rightarrow \quad \frac { { (t-2) }^{ 2 } }{ t-1 } <0\)
\(\Rightarrow \quad t-1<0\)
\(\Rightarrow \quad t<1\quad \Rightarrow \quad In\quad x<1\)
or x<e
but x>0
\(\therefore\) 0<x<3
\(\Rightarrow \quad x\epsilon (0,e)\)