Respuesta :

Hagrid
An equation with the form [tex]b^x=y[/tex] has a logarithmic form of [tex]log_b(y) = x.[/tex]. The logarithmic form of [tex]2^5=32[/tex] where in b = 2, x=5, and y=32 is
[tex]log_232=5[/tex].

Answer:

The answer is C. [tex]log_2(32)=5[/tex]

Step-by-step explanation:

The logarithmic form is:

[tex]log_b(x)=y\\b^y=x\\[/tex]

Where:

b is the base of the logarithmic.

In this case, the base could be "2" because the left side of the equation has a 2 powered by 5 and the right side of the equation has a 32 that is equal to 2 powered by 6. Also we can apply the next property:

[tex]log(a^b)=b*log(a)[/tex]

So, we have to reverse the form from exponential to logarithmic.

Then:

[tex]2^5=32\\log_2(2^5)=log_2(32)\\5*log_2(2)=log_2(32)\\5*1=log_2(32)\\log_2(32)=5[/tex]

Finally, the equation in logarithmic form is C. [tex]log_2(32)=5[/tex]

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