Question

find the logarithm of the number to the base 2 ​

Answer 1

Answer:

hope you get the solution

Answer 2

$\large\text{ \dfrac{1}{\sqrt[7]{8}} }$

We know that  $\dfrac{1}{x}=x^{-1}.$  By this, we write,

$\large\text{ \dfrac{1}{\sqrt[7]{8}}\ =\ \left(\sqrt[7]{8}\right)^{-1} }$

And we know that  $\large\text{ \sqrt[m]{a}=a^{\frac{1}{m}} }.$  Using this,

$\large\text{ \left(\sqrt[7]{8}\right)^{-1}\ =\ \left(8^\!^{\frac{1}{7}}\right)^{-1} }$

And, by using  $(a^m)^n=a^{mn},$

$\large\text{ \left(8^\!^{\frac{1}{7}}\right)^{-1}\ =\ 8^\!^{-\frac{1}{7}} }$

Now,

$\log_2\left(\dfrac{1}{\sqrt[7]{8}}\right)\ =\ \log_2\left(8^\!^{-\frac{1}{7}}\right)$

Taking  $8=2^3,$

$\large\text{ \log_2\left(8^\!^{-\frac{1}{7}}\right)\ =\ \log\left(\left(2^3\right)^\!^{-\frac{1}{7}}\right)\ =\ \log_2\left(2^\!^{-\frac{3}{7}}\right)=\bold{-\dfrac{3}{7}} }$

Hence  - 3/7  is the logarithm of  $\large\text{ \dfrac{1}{\sqrt[7]{8}} }.$

Let's check!

$\large\text{ 2^{^{-\frac{3}{7}}}\ =\ (2^{-3})^{^{\frac{1}{7}}}\ =\ \left(\dfrac{1}{8}\right)^{\frac{1}{7}}\ =\ \sqrt[7]{\dfrac{1}{8}}\ =\ \dfrac{1}{\sqrt[7]{8}} }$