Question 534847
<pre>
{{{32^(x+1)}}} = {{{64^(y+2)}}}

Notice that 32={{{2^5}}} and that 64 = {{{2^6}}}

Replacing these:


{{{(2^5)^(x+1)}}} = {{{(2^6)^(y+2)}}}

{{{2^(5(x+1))}}} = {{{2^(6(y+2))}}}

Now you can equate the exponents of 2 on each side:

5(x + 1) = 6(y + 2)

  5x + 5 = 6y + 12

  5x - 7 = y

f(x) = 5x - 7

Edwin</pre>