GATE General Aptitude - Logarithm
Exam Duration: 45 Mins Total Questions : 25
If logx \(\frac { 9 }{ 16 } =-\frac { 1 }{ 2 } ,\)then x is
- (a)
\(-\frac { 1 }{ 2 } \)
- (b)
\(\frac {2 }{3 } \)
- (c)
\(\frac { 81 }{ 256 } \)
- (d)
\(\frac { 256 }{81 } \)
logx\(\frac { 9 }{ 16 } \) =\(-\frac { 1 }{ 2 } \)
Hece, x-1/2=\(\frac { 9 }{ 12 } \)
\(\Rightarrow \quad \quad { x }^{ -1 }={ \left( \frac { 9 }{ 16 } \right) }^{ 2 }=\frac { 81 }{ 256 } \\ \Rightarrow \quad \quad x=\frac { 256 }{ 81 } \)
The value of 3 log 3 +2 log 2 is
- (a)
log 108
- (b)
log 106
- (c)
log 109
- (d)
None of these
3 log 3 +2 log =log 33 + log 22
=log 27 +log 4
=log 27 + log 4
=log(27\(\times\)4=log 108
The true statement of the following is
- (a)
log (xy)=log x log y
- (b)
log (xy) = log x + log y
- (c)
Botha (a) and (b)
- (d)
None of these
log(xy) = log x + log y
One is the value of
- (a)
log102
- (b)
log10 100
- (c)
log10 10
- (d)
log10 1000
logx x=1
The value of [log2 log2 log2 (65536) is
- (a)
8
- (b)
16
- (c)
4
- (d)
1
log2 log2 log2 log2 216= log2 log2 log2 (16)
=log2 log2 log2 (24)=log log (4)
=log2 log2 (22)= log2 (2) = 1
logx x is equal to
- (a)
\(\frac { X }{ { log }_{ e }Y } \)
- (b)
\(X{ log }_{ e }Y\)
- (c)
\(\frac { { log }_{ e }X }{ { log }_{ e }Y } \)
- (d)
\(\frac { { log }_{ e }Y }{ { log }_{ e }X } \)
\(\\ \\ \\ { log }_{ y }x=\frac { { log }_{ e }X }{ { log }_{ e }Y } \)
The value of \(\left( \frac { 1 }{ 3 } { log }_{ 10 }\quad 125-2{ log }_{ 10 }\quad 4+{ log }_{ 10 }\quad 32 \right) is\)
- (a)
\(\frac { 4 }{ 3 } \)
- (b)
3
- (c)
1
- (d)
7
As, log10 (125) 1/3 -log10 42+ log10 25
=log10 5 log1024+5log102
=log10\(\frac { 10 }{ 2 } \)-4log102 + 5log102
=log1010-log102-4log102+5log102=2
The value of logy X. logz Y . logX Z is
- (a)
log xyz
- (b)
xyz
- (c)
1
- (d)
0
\(\frac { logx }{ logy } \times \frac { logy }{ logz } \times \frac { logz }{ logx } =1\)
The value of log3 \((27\times \sqrt [ 4 ]{ 9 } \times \sqrt [ 3 ]{ 9 } )\quad is\)
- (a)
4
- (b)
\(4\frac { 1 }{ 3 } \)
- (c)
\(8\frac { 1 }{ 3 } \)
- (d)
\(4\frac { 1 }{ 6 } \)
Let log3 \((27\times \sqrt [ 4 ]{ 9 } \times \sqrt [ 3 ]{ 9 } )=x\)
\(\therefore\) 3x=\((27\times \sqrt [ 4 ]{ 9 } \times \sqrt [ 3 ]{ 9 } )=x\)
=33\(\times\)32/4\(\times\)32/3
\(\Rightarrow\) 3x=325/6\(\Rightarrow\)x=\(\frac { 25 }{ 6 } =4\frac { 1 }{ 6 } \)
If log4 x + log2 x -6 then the value of 'x' is
- (a)
16
- (b)
4
- (c)
2
- (d)
1
log4x + log2 x =6
\(Or\quad \quad \frac { logx }{ log4 } +\frac { logx }{ log2 } =6\\ \quad \quad \frac { logx }{ 2log2 } +\frac { logx }{ log2 } =6\Rightarrow 3logx=12logx\\ \quad \quad log\quad x=4\quad log\quad 2\\ \quad \quad logx=log16\\ \quad \quad x=16\)
log1/3 81 is equal to
- (a)
9
- (b)
27
- (c)
-4
- (d)
4
log1/3 81=x\(\Leftrightarrow \)(1-3)x =81=34
\(\Leftrightarrow { \left( \frac { 1 }{ 3 } \right) }^{ x }={ (3 })^{ 4 }\Leftrightarrow { \left( \frac { 1 }{ 3 } \right) }^{ x }={ \left( \frac { 1 }{ 3 } \right) }^{ -4 }\Rightarrow x=-4\)
Give log10 2= 0.3010, the value of log10 5 is
- (a)
0.6990
- (b)
0.6919
- (c)
0.6119
- (d)
0.7525
log105=log10\(\frac { 10 }{ 2 } \)=log10 10-log10 10-log102
=1-0.3010=-0.6990
If loga X=m, the value \({ log }_{ n }^{ 2 }\) X is
- (a)
\(-\frac { 1 }{ m } \)
- (b)
m
- (c)
m/2
- (d)
None of these
As, loga x=m\(\Rightarrow\)x=am
x=(a2)m/2
\(\therefore\) \({ log }_{ n }^{ 2 }\) x=m/2
If\(log=\frac { X }{ Y } =log\frac { Y }{ X } =log(X+Y),\quad then\)
- (a)
X+Y=1
- (b)
X-Y=0
- (c)
b=\(\frac { a-1 }{ a } \)
- (d)
a=b
\(Here,log=\frac { x }{ y } +log\frac { y }{ x } =log(x+y)\\ \quad \quad \quad \quad \Rightarrow log\frac { x }{ y } .\frac { y }{ x } =log(x+y)\\ \quad \quad \quad \quad \Rightarrow \quad log1==log(x+y)\Rightarrow =x+y=1\)
(log tan10 log tan 20 ...log tan 500) is
- (a)
1
- (b)
-1
- (c)
0
- (d)
\(\frac { 1 }{ \sqrt { 2 } } \)
( log tan10) (log tan 20) (log tan 30)
.................(log tan 460)......log tan 500)
=[(log tan 1)(log tan 2)
.........log(tan44) log(tan 460)
.......log tan 500]log tan 450
=[(log tan1).......(log tan 440)(logtan 460)
............log tan 500]\(\times\)0=0
The value \(\frac { 1 }{ 1+{ log }_{ x }(YZ) } +\frac { 1 }{ 1+{ log }_{ y }(XY) } +\frac { 1 }{ 1+{ log }_{ z }(XY) } is\)
- (a)
1
- (b)
\(\frac { 1 }{ { XY }^{ 2 } } \)
- (c)
X=Yz
- (d)
0
\(=\frac { 1 }{ { log }_{ x }(yx)+{ log }_{ x }x } +\frac { 1 }{ { log }_{ y }(xz)+{ log }_{ y }y } +\frac { 1 }{ { log }_{ z }(xy)+{ log }_{ z }xyz } \\ ={ log }_{ xyz }+{ log }_{ xyz }y+{ log }_{ xyz }xyz=1\)
Give that log103=0.3010, log103=0.4771 and log107=0.8491 then log10\(\frac { 108 }{ \sqrt { 7 } } \)is
- (a)
2.6123
- (b)
1.6088
- (c)
1.6320
- (d)
2.4558
log10\(\frac { 108 }{ \sqrt { 7 } } \)=log10108-log10\( \sqrt { 7 }\)
=log1023\(\times\)33-log1071/2
=2log102+3log103-\(\frac { 1 }{ 2 } \)log107
=2\(\times\)(0.3010)+3.(0.4245-\(\frac { 1 }{ 2 } \)(0.8491)
=0.6020+1.4313-0.4245=1.6088
If log(x+y)=log x=log y and x =11568, then yis equal to
- (a)
7.37776
- (b)
7
- (c)
5.3776
- (d)
5
log (x+y)=log x+log y
log (x+y)=logxy
(x+y)=xy
Or y=\(\frac { x }{ x-1 } =\frac { 1.1568 }{ 1.568-1 } =\frac { 1.1568 }{ 0.1568 } =7.37755=7.3776\)
Which is not correct?
- (a)
log10(1+2+3)=log10(1:2:3)
- (b)
log101=0
- (c)
log10 (2+3)=log10 2:3
- (d)
log1010=1
If a, b, c are three consecutive integers,, then log(ac+1) is equal to
- (a)
log(2b)
- (b)
x =2
- (c)
2 log b
- (d)
None of these
Let a=a
Then, b=a+1 and c=a+2
ac+1=a(a+2)+1
=a2+2a+1=(a+1)2
ac+1=b2
log(ac+1)= log b2=2log b
The solution of equation log7 [log4(X2)]=0 is
- (a)
x=1
- (b)
x=2
- (c)
x=\(\pm \)2
- (d)
x=-2
log7[log4(x2)-0=log71
\(\therefore\) log4(x2)=1=log44
x2=4
x=\(\pm \)2
If log r p=2, logr q=3, then the value of logp q is equal to
- (a)
1/3
- (b)
2/3
- (c)
3/2
- (d)
6
Here, logr P=2, logr q=3
By reaction, logp q=\(\frac { { log }_{ r }q }{ { log }_{ r }p } =\frac { 3 }{ 2 } \)
If log X2 Y2=a, and log\(\frac { X }{ Y } \)=b,then \(\frac { logX }{ logY } \)is equal to
- (a)
\(\frac { a-3b }{ a+2b } \)
- (b)
\(\frac { a+3b }{ a-2b } \)
- (c)
\(\frac { a+2b }{ a-3b } \)
- (d)
\(\frac { a-2b }{ a+3b } \)
log x2y2=a
2log x+2 log y =a
log x-log y=b
Solving log x =\(\frac { a+2b }{ 4 } \)
log y=\(\frac { a-3b }{ 4 } \)
\(\therefore\) \(\frac { log\quad x }{ log\quad y } =\frac { a-2b }{ a-3b } \)
If 22x+3=6x-1,equals to
- (a)
\(\frac { 4log\quad 2\quad +log3 }{ log3-log2 } \)
- (b)
\(\frac { 3log2+log3 }{ lo3-2log2 } \)
- (c)
\(\frac { log\quad 48 }{ log\quad 7 } \)
- (d)
None of these
22x+3=6x-1
(2x+3)log 2 =(x-1)log 6 = (x-1)(log2 + log 3)
2x loh 2 + log 2=x(log 2 + log 3)-log2-log3
Or x(log2-log3)=-4log2-log3
x=\(\frac { 4log2+log3 }{ log3-log2 } \)
The value of \({ log }_{ 3 }\left( 1+\frac { 1 }{ 3 } \right) +{ log }_{ 3 }\left( 1+\frac { 1 }{ 4 } \right) { log }_{ 3 }\left( 1+\frac { 1 }{ 5 } \right) +......+{ log }_{ 3 }\left( 1+\frac { 1 }{ 24 } \right) is\)
- (a)
-1+2log35
- (b)
2
- (c)
3
- (d)
4
\({ log }_{ 3 }\left( 1+\frac { 1 }{ 3 } \right) +{ log }_{ 3 }\left( 1+\frac { 1 }{ 4 } \right) { log }_{ 3 }\left( 1+\frac { 1 }{ 5 } \right) +......+{ log }_{ 3 }\left( 1+\frac { 1 }{ 24 } \right) \\ { log }_{ 3 }4-{ log }_{ 3 }3+{ log }_{ 3 }5-{ log }_{ 3 }4+{ log }_{ 3 }6-{ log }_{ 3 }5+......+{ log }_{ 3 }25-{ log }_{ 3 }24\\ =-{ log }_{ 3 }3+{ log }_{ 3 }25=-1+2{ log }_{ 3 }5\)