Question

Find the value of ${\log _{\sqrt 2 }}64.$

Answer 1

Hint:
First, we will use the formula ${\log _a}b = \dfrac{{\log a}}{{\log b}}$ on ${\log _{\sqrt 2 }}64$
Then, apply the property $\log {a^b} = b\log a$and solve further to get the answer.

Complete step by step solution:
Here, it is asked to find the value of ${\log _{\sqrt 2 }}64.$
Applying property ${\log _a}b = \dfrac{{\log a}}{{\log b}}$ on ${\log _{\sqrt 2 }}64$ we get,
$\Rightarrow {\log _{\sqrt 2 }}64 = \dfrac{{\log 64}}{{\log \sqrt 2 }}$
Also $64 = {2^6}$ and $\sqrt 2 = {2^{\dfrac{1}{2}}}$
$\Rightarrow{\log _{\sqrt 2 }}64 = \dfrac{{\log {2^6}}}{{\log {2^{\dfrac{1}{2}}}}}$
Now, applying property $\log {a^b} = b\log a$on ${\log _{\sqrt 2 }}64$, we get
\[{\log _{\sqrt 2 }}64 = \dfrac{{6\log 2}}{{\dfrac{1}{2}\log 2}}\]
$\Rightarrow {\log _{\sqrt 2 }}64 = \dfrac{6}{{\dfrac{1}{2}}}$
$\Rightarrow{\log _{\sqrt 2 }}64 = 6 \times 2$

$\therefore $${\log _{\sqrt 2 }}64 = 12$

Note:
There are different types of properties of logarithms which are used to find the solution of the particular questions which are stated below:
1) ${\log _b}\left( {x.y} \right) = {\log _b}\left( x \right) + {\log _b}\left( y \right)$
2) ${\log _b}\left( {\dfrac{x}{y}} \right) = {\log _b}\left( x \right) - {\log _b}\left( y \right)$
3) ${\log _b}\left( {{x^y}} \right) = y.{\log _b}\left( x \right)$
4) ${\log _b}\left( c \right) = \dfrac{1}{{{{\log }_c}\left( b \right)}}$
5) ${\log _b}\left( x \right) = \dfrac{{{{\log }_c}\left( x \right)}}{{{{\log }_c}\left( b \right)}}$
6) ${\log _b}\left( 0 \right)$= undefined
7) ${\log _b}\left( 1 \right) = 0$
8) ${\log _b}\left( b \right) = 1$