Question 855564
<pre>
To prove: {{{log(8,59)=log(10,59)/log(10,8)}}}

Here goes:

Let {{{log(8,59)=x}}}

Change that log equation to its 
equivalent exponent form:

{{{8^x=59}}}

Now take logs base 10 of both sides:

{{{8^x=59}}}

{{{log(10,8^x)=log(10,59)}}}

Use the "jump-over" rule of logarithms where the exponent
"jumps over both its base and the word "log" and turns into a
multiplier. Like this:

{{{x*log(10,8)=log(10,59)}}}

See how the x jumped over the "8" and the "log" and became
a multiplier instead of being an exponent. 

Now we divide both sides by {{{log(10,8)}}} to solve for x.

{{{x=log(10,59)/log(10,8)}}}

And now we go back and remember what x was -- we let {{{log(8,59)=x}}}

So we have:

{{{log(8,59)=log(10,59)/log(10,8)}}}

and it's proved.

Edwin</pre>