AMCAT Logarithms Quiz-1

$\log_{6}\left [ \frac{1}{216} \right ]=n$
Then,$6^n = \frac{1}{216}$
$6^n = 6^{-3}\rightarrow n=-3$
= -3

$\log_{25}x+\log_{5}x=3$
$\frac{1}{2}\log_{5}x+\log_{5}x=3$
$\log_{5}x^{\frac{3}{2}}=3$
$3\times \log_{5}5$
$x^{3/2}=\log_{5}5^3$
$x= 5^{3\times\frac{2}{3}}$
x = 25

Let $\log_{5}125=x$
Then, $5^x=125=5^3\rightarrow x=3$
Let $\log_{2}(log_{5}125)=x$
Then$\log_{2}(3)=x$ or $2^x=3\rightarrow x=\log_{2}(3)$
$\log_{2}(log_{5}125)=\log_{2}3$

$\log_{3}(243)=n$
therefore $3^n=3^5\rightarrow n=5$
$\log_{3}(243)=5$

Since $\log4^{30}=30\log(4)$
$30[log2^2]=2\times30\log2=60\times0.3010=18.06$

$\log3^6$ ⇒ 6 log 3 = 6 × 0.4771 ⇒ 2.8626= 2.8

$\log_{7}(49)+\log_{7}(7)=x$ ⇒ $\log_{7}(49\times7)=x$

$x=\log_{7}(343)=\log_{7}(7^3)=3\log_{7}7\rightarrow 3\times1=3$

$\log_{4096}x=-\frac{1}{4}$
Then,
$x=(4096)^{\frac{-1}{4}}\rightarrow (8^4)^{-1/4}=\frac{1}{8}$

$\log_{10}40\rightarrow \log_{10}10\times4= \log_{10}10+\log_{10}4$
$1+2\log_{10}2$
= 1 + 2( 0.3010)
= 1 + 0.6020
= 1.6020

$\log_{4}x=-10\rightarrow x=4^{-10}=\frac{1}{4^{10}}=\frac{1}{1048576}$