Question

Given the exponential equation 2x = 128, what is the logarithmic form of the equation in base 10?

Answer 1

$2^{x} = 128$
$log_2 128 = x$
$x = \frac{log_{10}128}{log_{10}2}$

Answer 2

Answer:

$log_{2}128= \frac{log_{10} 128}{log_{10} 2}$.  

Step-by-step explanation:

Given : $2^{x}$ = 128.

To find :  what is the logarithmic form of the equation in base 10

Solution : We have given that $2^{x}$ = 128.

By th change base rule ( inverse of exponential function )

$a^{b}$ = c is equal to  $log_{a}(c)$ = b

Then $2^{x}$ = 128 in to logarithm form  $log_{2}(128)$ = x.

Then in to the base 10 logarithmic form.

By the change of base formula $log_{b}a= \frac{log_{10} a}{log_{10} b}$.

Then , $log_{2}128= \frac{log_{10} 128}{log_{10} 2}$.  

Therefore, $log_{2}128= \frac{log_{10} 128}{log_{10} 2}$.