Respuesta :

[tex]2^{x} = 128[/tex]
[tex]log_2 128 = x[/tex]
[tex]x = \frac{log_{10}128}{log_{10}2}[/tex]

Answer:

[tex]log_{2}128= \frac{log_{10} 128}{log_{10} 2}[/tex].  

Step-by-step explanation:

Given : [tex]2^{x}[/tex] = 128.

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

Solution : We have given that [tex]2^{x}[/tex] = 128.

By th change base rule ( inverse of exponential function )

[tex]a^{b}[/tex] = c is equal to  [tex]log_{a}(c)[/tex] = b

Then [tex]2^{x}[/tex] = 128 in to logarithm form  [tex]log_{2}(128)[/tex] = x.

Then in to the base 10 logarithmic form.

By the change of base formula [tex]log_{b}a= \frac{log_{10} a}{log_{10} b}[/tex].

Then , [tex]log_{2}128= \frac{log_{10} 128}{log_{10} 2}[/tex].  

Therefore, [tex]log_{2}128= \frac{log_{10} 128}{log_{10} 2}[/tex].