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}