Logarithms Questions for XAT (PDF)
Download important Logarithms Questions for XAT PDF based on previously asked questions in XAT exam. Practice Logarithms Questions PDF for XAT exam.
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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: Let $u = ({\log_2 x})^2 – 6 {\log_2 x} + 12$ where x is a real number. Then the equation $x^u = 256$, has
a) no solution for x
b) exactly one solution for x
c) exactly two distinct solutions for x
d) exactly three distinct solutions for x
Question 3: 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 4: If x >= y and y > 1, then the value of the expression $log_x (x/y) + log_y (y/x)$ can never be
a) -1
b) -0.5
c) 0
d) 1
Question 5: If $\log_{2}{\log_{7}{(x^2 – x+37)}}$ = 1, then what could be the value of ‘x’?
a) 3
b) 5
c) 4
d) None of these
Question 6: 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 7: What is the value of $\sqrt{\frac{a}{b}}$, If $\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+1$
a) -5/3
b) 2
c) 5/3
d) 1
Question 8: 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 9: If $\log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression, then x is equal to
a) $\frac{8}{3}$
b) $\log_{3}{8}$
c) $\log_{2}{3}$
d) $8$
Question 10: 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
XAT Decision making practice questions
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 (B)
$x^u = 256$
Taking log to the base 2 on both the sides,
$u * \log_{2}{x} = \log_{2}{256}$
=>$[({\log_2 x})^2 – 6 {\log_2 x} + 12] * \log_{2}{x} = 8$
$(log_2 x)^3 – 6(log_2 x)^2 + 12log_2 x = 8$
Let $log_2 x = t$
$t^3 – 6t^2 +12t – 8 = 0$
$(t-2)^3 = 0$
Therefore, $log_2 x = 2$
=> $x = 4$ is the only solution
Hence, option B is the correct answer.
3) 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
4) Answer (D)
$log_x (x/y) + log_y (y/x)$ = $1 – log_x (y) + 1 – log_y (x)$
= $2 – (log_x y + 1/log_x y)$ <= 0 (Since $log_x y + 1/log_x y$ >= 2)
So, the value of the expression cannot be 1.
5) Answer (C)
$\log_{2}{\log_{7}{(x^2 – x+37)}}$ = 1
$\log_{7}{(x^2 – x+37)}$ = $2$
$(x^2 – x+37)$ = $7^{2}$
Given eq. can be reduced to $x^2 – x + 37 = 49$
So x can be either -3 or 4.
6) 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’
7) Answer (C)
$\sqrt{\frac{a}{b}}$, If $\log_{4}\log_{4}4^{a-b}=2\log_{4}(\sqrt{a}-\sqrt{b})+\log_{4}{4}$
i.e. $\log_{4}\log_{4}4^{a-b}=\log_{4}((\sqrt{a}-\sqrt{b})^2)*4$
i.e. $\log_{4}4^{a-b}=((\sqrt{a}-\sqrt{b})^2)*4$
i.e. (a-b)*$\log_{4}4=((\sqrt{a}-\sqrt{b})^2)*4$
i.e. a-b = 4a+4b-8$\sqrt{ab}$
i.e. 3a + 5b – 8$\sqrt{ab}$ = 0
i.e. $3\sqrt\frac{a}{b}^2$ – 8$\sqrt\frac{a}{b}$+5 = 0
put $\sqrt\frac{a}{b}$ = t
therefore 3$t^2$ – 8t + 5 = 0
solving we get t = 1 or t = $\frac{5}{3}$
i.e. $\sqrt\frac{a}{b}$ = 1 or $\frac{5}{3}$
but if $\sqrt\frac{a}{b}$ = 1 then a=b then $\log_{4}(\sqrt{a}-\sqrt{b})$ will become indefinite
Therefore $\sqrt\frac{a}{b}$ = $\frac{5}{3}$
Therefore our answer is option ‘C’
8) 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.
9) Answer (B)
If $log{3}, log(3^{x} – 2)$ and $log (3^{x}+ 4)$ are in arithmetic progression
Then, $2*log(3^{x} – 2) = log{3}+log (3^{x}+ 4)$
Thus, $log{(3^{x} – 2)^2} = log{3(3^x+4)}$
Thus, $(3^{x} – 2)^2 = 3(3^x+4)$
=> $3^{2x} – 4*3^x +4 = 3*3^x + 12$
=> $3^{2x} – 7*3^x – 8 = 0$
=> $(3^x+1)*(3^x-8) = 0$
But $3^x+1 \neq 0$
Thus, $3^x = 8$
Hence, $x = log_{3}{8}$
Hence, option B is the correct answer.
10) 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, => $\dfrac{1}{9}*(\log{x})^2 = (\log{10})^2=1$
Thus, $(\log{x})^2 = 9$
Thus $\log x = 3$ or $-3$
Thus, $ x = 1000$ or $\dfrac{1}{1000}$
From amongst the given options, 1000 is the correct answer.
Hence, option D is the correct answer.
We hope this Logarithms Questions PDF for XAT with Solutions will be helpful to you.
