Question 895153

Dear Sir/Madam, 


Please help me with this logarithm problem, 

If {{{b = log(3,x)}}}, find x in {{{log(b,(log(3,x^2)))=2}}}



note: the base in log3x is 3, the base in logb is b and the base in log3x^2 is 3


Thank you very much!
<pre>
{{{log(b,(log(3, x^2))) = 2}}}
{{{log(3, x^2)}}} = {{{2log (3, x)}}}
Since {{{log (3, x) = b}}}, then {{{2log (3, x)}}}, or {{{log (3, x^2)}}} = 2b
{{{log(b, 2b) = 2}}} ------ Substituting 2b for {{{log (3, x^2)}}} in {{{log(b,(log(3, x^2))) = 2}}}
{{{b^2 = 2b}}}
{{{b^2 - 2b = 0}}}
{{{b(b - 2) = 0}}}
Thus, b = 2                 OR            b = 0 (ignore)

Now, since {{{b = log(3, x)}}}, and b = 2, then:
{{{2 = log (3, x)}}}
{{{x = 3^2}}}, or {{{highlight_green(x = 9)}}}