<PRE><B>Question 1027799</B>
&lt;p&gt;Have you tried taking the logarithm of both sides? If you don&#39;t know what base to use, don&#39;t choose one, just use {{{log_b(x)}}} to mean the logarithm of {{{x}}} to base {{{b}}}. You will see later that the base you choose is irrelevant.&lt;/p&gt;

&lt;p&gt;We will need a couple of properties of logarithms in order to fully solve this, but the most important one is this:
{{{ a*log_b(c) = log_b(c^a) }}}
If we apply this rule to remove the exponent on the lhs we get {{{ (a+2)log_b(7) }}}, similarly the rhs is {{{ 2a* log_b(9) }}}.&lt;/p&gt;

&lt;p&gt;The resulting equation is actually linear in a, so you should be able to rearrange the equation so that it looks like {{{ a*L_1 = L_2 }}} where {{{ L_1, L_2 }}} are expressions in terms of logarithms to base b of numbers.&lt;/p&gt;

&lt;p&gt;You can then get an exact expression for a, {{{ a=L_2/L_1 }}}.&lt;/p&gt;

&lt;p&gt;This expression still has a {{{b}}} in it, but if you try with different values of {{{b}}} you should find you always get the same answer.&lt;/p&gt;

&lt;p&gt;See if you can use some of the other properties of logarithms to show that the results will always be the same regardless for what value of {{{b&gt;0}}} that you choose.&lt;/p&gt;

</PRE>