Logarithms Questions for IIFT PDF
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Question 1: If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7/2)$ are in arithmetic progression, then the value of x is equal to
a) 5
b) 4
c) 2
d) 3
Question 2: If $log_y x = (a*log_z y) = (b*log_x z) = ab$, then which of the following pairs of values for (a, b) is not possible?
a) (-2, 1/2)
b) (1,1)
c) (0.4, 2.5)
d) ($\pi$, 1/ $\pi$)
e) (2,2)
Question 3: If $f(x) = \log \frac{(1+x)}{(1-x)}$, then f(x) + f(y) is
a) $f(x+y)$
b) $f{\frac{(x+y)}{(1+xy)}}$
c) $(x+y)f{\frac{1}{(1+xy)}}$
d) $\frac{f(x)+f(y)}{(1+xy)}$
Question 4: If $\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$. Then x is
a) 2
b) 4
c) 16
d) 12
Question 5: Find the value of x from the following equation:
$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+1$
a) 2/7
b) 7/2
c) 9/2
d) None of the above
Question 6: If $log_{10} x – log_{10} \sqrt[3]{x} = 6log_{x}10$ then the value of x is
a) 10
b) 30
c) 100
d) 1000
Question 7: $(1+5)\log_{e}3+\frac{(1+5^{2})}{2!}(\log_{e}3)^{2}+\frac{(1+5^{3})}{3!}(\log_{e}3)^{3}+…$
a) 12
b) 244
c) 243
d) 245
Question 8: If $log(2^{a}\times3^{b}\times5^{c} )$is the arithmetic mean of $log ( 2^{2}\times3^{3}\times5)$, $log(2^{6}\times3\times5^{7} )$, and $log(2 \times3^{2}\times5^{4} )$, then a equals
Question 9: If $\log_{2}({5+\log_{3}{a}})=3$ and $\log_{5}({4a+12+\log_{2}{b}})=3$, then a + b is equal to
a) 59
b) 40
c) 32
d) 67
Question 10: $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$=?
a) $\frac{1}{2}$
b) 10
c) 0
d) −4
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Answers & Solutions:
1) Answer (D)
$2 log (2^x – 5) = log 2 + log (2^x – 7/2)$
Let $2^x = t$
=> $(t-5)^2 = 2(t-7/2)$
=> $t^2 + 25 – 10t = 2t – 7$
=> $t^2 – 12t + 32 = 0$
=> t = 8, 4
Therefore, x = 2 or 3, but $2^x$ > 5, so x = 3
2) Answer (E)
$log_y x = ab$
$a*log_z y = ab$ => $log_z y = b$
$b*log_x z = ab$ => $log_x z = a$
$log_y x$ = $log_z y * log_x z$ => $log x/log y$ = $log y/log z * log z/log x$
=> $\frac{log x}{log y} = \frac{log y}{log x}$
=> $(log x)^2 = (log y)^2$
=> $log x = log y$ or $log x = -log y$
So, x = y or x = 1/y
So, ab = 1 or -1
Option 5) is not possible
3) Answer (B)
If $f(x) = \log \frac{(1+x)}{(1-x)}$ then $f(y) = \log \frac{(1+y)}{(1-y)}$
Also Log (A*B)= Log A + Log B
f(x)+f(y) = $ \log \frac{(1+x)(1+y)}{(1-x)(1-y)}$ solving we get $\log { \frac{1+ \frac{(x+y)}{(1+xy)}}{1- \frac{(x+y)}{(1+xy)}}}$
Hence option B.
4) Answer (B)
$\log_{2}{x}.\log_{\frac{x}{64}}{2}=\log_{\frac{x}{16}}{2}$
i.e. $\frac{log{x}}{log{2}} * \frac{log_{2}}{log{x}-log{64}} = \frac{log{2}}{log{x}-log{16}}$
i.e. $\frac{log{x} * (log{x}-log{16})}{log{x}-log{64}}$ = $\log{2}$
let t = log x
Therefore, $\frac{t * (t-log{16})}{t-log{64}}$ = $\log{2}$
$t^2-4*log 2*t = t*log 2-6*(log 2)^2$
I.e. $t^2-5*log 2*t-6*(log 2)^2$ = 0
I.e. $t^2-3*log 2*t-2*log 2*t-6*(log 2)^2$ = 0
i.e. $t*(t-3*log 2)-2*log 2*(t-3*log 2)$ = 0
i.e $t=2*log 2$ or $t=3*log 2$
i.e $log x=log 4$ or $log x=log 8$
therefore $x=4$ or $8$
therefore our answer is option ‘B’
5) Answer (B)
$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+1$ can be written as
$\log_{10}{3}+\log_{10}(4x+1)=\log_{10}(x+1)+\log_{10}{10}$
We know that $\log_{10}{a}+\log_{10}{b}=\log_{10}{ab}$
$\log_{10}{3*(4x+1)}=\log_{10}{(x+1)*10}$
$12x+3=10x+10$
$x=7/2$. Hence, option B is the correct answer.
6) Answer (D)
$\log_{10} x – \log_{10} \sqrt[3]{x} = 6\log_{x}10$
Thus, $\dfrac{\log {x}}{\log {10}}$ – $\dfrac{1}{3}*\dfrac{\log {x}}{\log {10}}$ = $6*\dfrac{\log{10}}{\log{x}}$
=> $\dfrac{2}{3}*\dfrac{\log {x}}{\log {10}}$ = $6*\dfrac{\log{10}}{\log{x}}$
Thus, => $\dfrac{1}{9}*(\log{x})^2 = (\log{10})^2$
Thus, $ x = 1000$
Hence, option D is the correct answer.
7) Answer (B)
Splitting the above mentioned series into two series
A = $\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+…$
B = $5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+…$
We know that $e^{x}$ =$1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…$
So $e^{x}-1$ = $x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+…$
On solving two series A and B
A = $\log_{e}3+\frac{1}{2!}(\log_{e}3)^{2}+\frac{1}{3!}(\log_{e}3)^{3}+…$ =$e^{\log_{e}3}-1$ = $3-1$ =$2$
B = $5\log_{e}3+\frac{5^{2}}{2!}(\log_{e}3)^{2}+\frac{5^{3}}{3!}(\log_{e}3)^{3}+…$=$e^{\log_{e}3^{5}}-1$=$3^{5}-1$=$242$
A+B = $2 + 242$ = $244$
8) Answer: 3
$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3} $
$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3} $
$log(2^{a}\times3^{b}\times5^{c} )$ = $ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $
$3log(2^{a}\times3^{b}\times5^{c} )$ = $ log ( 2^{9}\times3^{6}\times5^{12}) $
Hence, 3a = 9 or a = 3
9) Answer (A)
$\log_{2}({5+\log_{3}{a}})=3$
=>$5 + \log_{3}{a}$ = 8
=>$ \log_{3}{a}$ = 3
or $a$ = 27
$\log_{5}({4a+12+\log_{2}{b}})=3$
=>$4a+12+\log_{2}{b}$ = 125
Putting $a$ = 27, we get
$\log_{2}{b}$ = 5
or, $b$ = 32
So, $a + b$ = 27 + 32 = 59
Hence, option A is the correct answer.
10) Answer (A)
We know that $\dfrac{1}{log_{a}{b}}$ = $\dfrac{log_{x}{a}}{log_{x}{b}}$
Therefore, we can say that $\dfrac{1}{log_{2}{100}}$ = $\dfrac{log_{10}{2}}{log_{10}{100}}$
$\Rightarrow$ $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$
$\Rightarrow$ $\dfrac{log_{10}{2}}{log_{10}{100}}$-$\dfrac{log_{10}{4}}{log_{10}{100}}$+$\dfrac{log_{10}{5}}{log_{10}{100}}$-$\dfrac{log_{10}{10}}{log_{10}{100}}$+$\dfrac{log_{10}{20}}{log_{10}{100}}$-$\dfrac{log_{10}{25}}{log_{10}{100}}$+$\dfrac{log_{10}{50}}{log_{10}{100}}$
We know that $log_{10}{100}=2$
$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$
$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{\dfrac{2*5*20*50}{4*10*25}}]$
$\Rightarrow$ $\dfrac{1}{2}*[log_{10}10]$
$\Rightarrow$ $\dfrac{1}{2}$
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