Practice Logarithm - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) What is the solution of the equation x $log_{10} ({10}/3) + log_{10} 3 = log_{10} (2 + 3^x)$ + x ?
(a)
(b)
(c)
(d)
$x log_{10}({10}/3) + log_{10} 3 = log_{10} (2 + 3^x) + x$
$xlog_{10} 10 - x log_{10} 3 + log_{10} 3 = log_{10} (2 + 3^x)$ + x
x - $log_{10^3^x} + log_{10^3} = log_{10} (2 + 3^x)$ + x
$log_{10[3/{3^x}]} log_{10} (2 + 3^x)$
⇒$3^{1–x} = 2 + 3^x$
⇒$3^{1–x} –3^x = 3^1 – 3^0$
x = 0
Q-2) What is the value of $2 log(5/8) + log({128}/{125}) + log (5/2)?$
(a)
(b)
(c)
(d)
$2 log (5/8) + log ({128}/{125}) + log (5/2)$
= $log(5/8)^2 + log({128}/{125}) + log(5/2)$
= $log {5^2 × 128 × 5}/{8^2 × 125 × 2} = log {5^2 × 2^7 × 5}/{(2^3)^2 × 5^3 × 2}$
= $log {2^7 × 5^3}/{2^6 × 5^3 × 2} = log {2^7 × 5^3}/{2^7 × 5^3}$ = log 1 = 0
Q-3) What is the value of $({1/3} \text"log"_{10} 125 -2\text"log"_{10} 4 + \text"log"_{10} 32 + \text"log"_{10} 1)$
(a)
(b)
(c)
(d)
(e)
${1/3} \text"log"_{10} 125 – 2{\text"log"_{10}} 4 + {\text"log"_{10}} 32 + {\text"log"_{10}}$ 1
=${1/3} \text"log"_{10} (5)^3 – 2{\text"log"_{10}}(2)^2 + {\text"log"_{10}}(2)^5$ + 0
=$\text"log"_{10} 5 – 4{\text"log"_{10}} 2 + 5{\text"log"_{10}2}$
= $\text"log"_{10}5 +\text"log"_{10}2 = \text"log"_{10}10 = 1$
Q-4) If ${\text"log"_7}{\text"log"_5} (√x + 5 + √x)$ = 0 , find the value of x.
(a)
(b)
(c)
(d)
(e)
$\text"log"_7 \text"log"_5(√x + 5 + √x)$ = 0
use $\text"log"_a$ x = b⇒$a^b$=x
∴ $\text"log"_5 (√x + 5 + √x) = 7^0$ = 1
$√x + 5 + √ x = 5^1 = 5⇒2√x$ = 0
∴ x = 0
Q-5) What is the value of ${[log_{13}(10)]}/{[log_{169}(10)]}$
(a)
(b)
(c)
(d)
${log_{13}(10)}/{log_{169}(10)} = {log_{13}(10)}/{log_{13^2}(10)}$
$(∵ log_{a^b} c = 1/b log_a c)$
= ${log_{13}10}/{1/2 log_{13}10} = 1/{1/2}$ = 2
Q-6) If ${\text"log"_a} b = 1/2, {\text"log"_b}$ c = $1/3$ and ${\text"log"_c}$ a = $k/5$, then the value of k is
(a)
(b)
(c)
(d)
(e)
${\text"log"_a} b = {1/2}, {\text"log"_b}c= {1/3}$, and ${\text"log"_c} a = k/5$
⇒${\text"log" b}/{\text"log" a} = {1/2}, {\text"log" c}/{\text"log" b} = {1/3}, {\text"log" a}/{\text"log" c} = k/5$
⇒${1/2} × {1/3} × {k/5}$ = 1⇒k = 30
Q-7) If ${\text"log"_8} x + {\text"log"_8} {1/6} = 1/3$, then find the value of x.
(a)
(b)
(c)
(d)
(e)
${\text"log"_8}x +{ \text"log"_8} {1/6} = 1/3$
or ${\text"log"_8} (x × {1/6}) = 1/3 or {\text"log"_8} (x/6) = 1/3$
or $x/6 = (8)^{1/3}$ {∵ $\text"log"_a$ b = x⇔(a)x=b}
or ${x/6} = (2^3)^{1/3}$ or x = 12
Q-8) It is given that $log_{10} 2$ = 0.301 and $log_{10} 3 = 0.477.$ How many digits are there in $(108)^{10}$ ?
(a)
(b)
(c)
(d)
$log (108)^{10} = 10 log 108 = 10 log (2^2 × 3^3)$ = 10 (2log2 + 3log3)
= 10 (2 × 0.301 + 3 × 0.477 ) = 10 (.602 + 1.431)
= 10 × 2.033 = 20.33
integral part = 20
No. of digits = 20 + 1 = 21
Q-9) There are n zeroes appearing immediately after the decimal point in the value of $(0.2)^{25}$. It is given that the value of $log_{10}2$ = 0.30103. The value of n is
(a)
(b)
(c)
(d)
Let y = $(0.2)^{25}$
Talking log on both sides, we got
log(y) = $(0.2)^{25}$ = 25. log(0.2)
= 25 $(log (2/{10}))$ = 25 (log2 – log10)
= 25 (0.330103 – 1) = –17.47475
Hence, n = 17
Q-10) If $\text"log" 2$ = 0.30103, then find the number of digits in $2^{56}$ .
(a)
(b)
(c)
(d)
(e)
Let x = $2^{56}$
⇒$\text"log" x = 56 \text"log" 2$ = 56 × 0.30103 = 16.85
∴ Number of digits in $2^{56}$ = 17
Q-11) What is the logarithm of 0.0001 with respect to base 10?
(a)
(b)
(c)
(d)
Let $log_{10}$ 0.0001 = a
a = $log_{10} {1/{(10)^4}$
= $log_{10} 1 – log_{10} (10)^4$ = 0 – 4 = –4
Q-12) What is the number of digits in $2^{40}$ ? (Given that $log_{10} 2$ = 0.301)
(a)
(b)
(c)
(d)
let $2^{40} = 10^x$
log $2^{40} = log 10^x$
40 $log^2 = x log^{10}$
40 (.301) = x
12.04 = x
i.e. x > 12
∴ x = 13
No. of digit in $2^{40}$ is 13
Q-13) If a = $\text"log"_{24} 12, b = \text"log"_{36}$ 24, C = $\text"log"_{48}$ 36. Then 1 + abc is equal to
(a)
(b)
(c)
(d)
(e)
abc = ${\text"log" 12}/{\text"log" 24} . {\text"log" 24}/{\text"log" 36} . {\text"log" 36}/{\text"log" 48} = {\text"log" 12}/{\text"log" 48}$
∴ 1 + abc = ${\text"log" 48 + \text"log"12}/{\text"log" 48} = {\text"log" (48.12)}/{\text"log" 48}$
= ${\text"log"24}^2/{\text"log" 48} = 2.{\text"log" 24}/{\text"log" 48}$ = 2bc
Q-14) $\text"log" 216 √6$ to the base 6 is
(a)
(b)
(c)
(d)
(e)
$\text"log"_{6} 216 √6 = \text"log"_6 (6)^3 (6)^{1/2}$
= $\text"log"_6 (6)^{7/2} = {7/2}{ \text"log"_6} 6 = 7/2$ (∵ $\text"log"_a$ a = 1)
Q-15) What is the solution of $log_{10} [1- [1 - (1 - x^2)^{-1}]^{-1}]^{- 1/2}$ = 1 ?
(a)
(b)
(c)
(d)
$log_{10} [1 - [1 - (1 - x^2)^{-1}]^{-1}]^{-1/2}$ = 1
⇒$log_{10}[1 - [1 - 1/{1 - x^2}]^{-1}]^{-1/2}$ = 1
⇒$log_{10}[1 - [{- x^2}/{1 - x^2}]^{-1}]^{- 1/2}$ = 1
⇒$log_{10}[1 - {(1 - x^2)}/{- x^2}]^{-1/2}$ = 1
⇒$log_{10}[{-x^2 - (1 - x^2)}/{- x^2}]^{-1/2}$ = 1
⇒$log_{10}[1/{x^2}]^{-1/2}$ = 1
⇒$log_{10} x = log_{10}10$
x = 10
⇒$log_{10}$
Q-16) Find $\text"log"_5. ^3√5$
(a)
(b)
(c)
(d)
(e)
Let m = ${\text"log"_5}^3√5$, then $5^m = ^3√5 = (5)^{1/3}$
So, m = $1/3$. Thus, ${\text"log"_5} ^3√5 = 1/3$ .
Q-17) If log x = 1.2500 and y = $x^{logx}$ , then what is log y equal to?
(a)
(b)
(c)
(d)
We have, y = $x^{logx}$
Taking log on both sides log
y = log $(x^{logx})$
= logx. logx = $(1.25)^2$ = 1.5625
Q-18) What is the value of $(log_{1/2}2)(log_{1/3}3)(log_{1/4}4).......(log_{1/{1000}}1000)$ ?
(a)
(b)
(c)
(d)
$(log_{1/2}2)(log_{1/3}3)(log_{1/4}4).......(log_{1/{1000}}1000)$
= $({log 2}/{log_{1/2}})({log 3}/{log_{1/3}})({log 4}/{log_{1/4}})......({log 1000}/{log_(1/{1000})})$
$(∵ log_b a = {log a}/{log b})$
= $({log 2}/{- log 2})({log 3}/{- log 3})({log 4}/{- log 4}).........({log 1000}/{- log 1000})$
= (-1) × (-1) × (-1) × ..... × (-1)
(∵ number of factors is odd)
= –1
Q-19) Let XYZ be an equilateral triangle in which XY = 7 cm. If A denotes the area of the triangle, then what is the value of $log_{10}A^4$ ? (Given that $log_{10}1050 = 3.0212$ and $log_{10}35$ = 1.5441)
(a)
(b)
(c)
(d)
$log_{10} 1050 = log_{10}$(3 × 10 × 35)
= $log_{10} 3 + 1 + log_{10} 35$
⇒ 3.0212 = $log_{10} 3 + 1 + 1.5441$
⇒$log_{10} 3$ = 0.4771
Now, $log_{10} 35 = log {7 × 10}/2$
= $log_{10} 7 + log_{10} - log_{10} 2$
⇒ $log_{10}7$ = 0.8451
Now, A = ${√3}/4 × (7)^2$
$log{10^A^4} = 4 log_{10}A$
= 4 log ${√3 × 7^2}/4$
= $4[1/2 log_3 + 2 log_{10} 7 - 2 log 2]$ = 5.3070
Q-20) What is the value of $(\text"log"_{1/2} 2) (\text"log"_{1/3}3) (\text"log"_{1/4}4).....(\text"log"_{1/{1000}}1000)$
(a)
(b)
(c)
(d)
(e)
$(\text"log"_{1/2}2)(\text"log"_{1/3}3)(\text"log"_{1/4}4).......(\text"log"_{1/{1000}}1000)$
=$({\text"log" 2}/{\text"log"{1/2}})({\text"log" 3}/{\text"log"{1/3}})({\text"log" 4}/{\text"log"{1/4}})......({\text"log" 1000}/{\text"log"{1/{1000}}})$ $(∵ \text"log"_b a = {\text"log" a}/{\text"log" b})$
= $({\text"log" 2}/{- \text"log" 2})({\text"log" 3}/{-\text"log" 3})({\text"log" 4}/{-\text"log" 4}) ...... ({\text"log" 1000}/{\text"log" 1000})$
= (-1) × (-1) × (-1) × .....× (-1)
(∵ number of terms is odd) = -1