Question 317476
<pre><b>
{{{2^(5x)*16^(1-x)=4^(x-3)}}}

Write 16 as {{{(2^4)}}} and write 4 as {{{(2^2)}}}

{{{2^(5x)*(2^4)^(1-x)=(2^2)^(x-3)}}}

Remove the parentheses by multiplying the inner exponents by
the outer exponents:

{{{2^(5x)*2^(4*(1-x))=2^(2*(x-3))}}}

Distribute in the exponents:

{{{2^(5x)*2^(4-4x)=2^(2x-6)}}}

Add the exponents of 2 on the left:

{{{2^(5x+4-4x)=2^(2x-6)}}}

{{{2^(x+4)=2^(2x-6)}}}

The bases are the same and are positive and not 1, so
we may set the exponent of 2 on the left equal the exponent
of 2 on the right:

{{{x+4=2x-6}}}
{{{10=x}}}

Edwin</pre>