Question 189335
Prove that 

{{{log(B,a^x) = x*log(B,a)}}}
<pre><font size = 4 color = "indigo"><B>
Let {{{M}}} equal the left side

{{{M = log(B,a^x)}}} 

Let {{{N}}} equal the right side

{{{N = x*log(B,a)}}}

Take the first equation:

{{{M = log(B,a^x)}}}

By definition of logarithms,

{{{B^M = a^x}}}

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Take the second equation:

{{{N = x*log(B,a)}}}

Divide both sides by x

{{{ N/x = log(B,a) }}}

By definition of logarithms,

{{{B^(N/x)=a}}}

Raise both sides to the x power

{{{(B^(N/x))^x=a^x}}}

{{{B^N=a^x}}}

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So we have the two equations:

{{{B^M = a^x}}}
{{{B^N=a^x}}}

Therefore

{{{B^M=B^N}}} since both equal to {{{a^x}}}

Therefore if B is positive and not equal to 1. then

{{{M=N}}}

Therefore

{{{log(B,a^x) = x*log(B,a)}}}

Edwin</pre>