Question 1144759
<pre>
{{{system(2^m*(1/8)^n = 128, 4^m/2^(-4n) = 1/6)}}}

{{{system( 2^m*(8^(-1))^n = 2^7, 4^m*2^(4n) = 1/6)}}}

{{{system(2^m*8^(-n) = 2^7, (2^2)^m*2^(4n) = 1/6)}}}

{{{system(2^m*(2^3)^(-n) = 2^7, 2^(2m)*2^(4n) = 1/6)}}}

{{{system(2^m*2^(-3n) = 2^7, 2^(2m+4n) = 1/6)}}}

{{{system(2^(m-3n) = 2^7, 2^(2m+4n) = 1/6)}}}

{{{system(m-3n = 7, 2^(2m+4n) = 1/6)}}}

{{{system(m=3n+7, 2^(2m+4n) = 1/6)}}}

By substitution:

{{{2^(2(3n+7)+4n) = 1/6}}}

{{{2^(10n+14) = 1/6}}}

Multiply both sides by 2

{{{2*2^(10n+14) = 2*expr(1/6)}}}

{{{2^(10n+14+1) = 1/3}}}

{{{2^(10n+15) = 3^(-1)}}}

Take natural logs of both sides:

{{{ln(2^(10n+15)) = ln(3^(-1))}}}

{{{(10n+15)ln(2) = -ln(3)}}}

Divide both sides by ln(2)

{{{10n+15=-ln(3)/ln(2)}}}

{{{10n=-15-ln(3)/ln(2)}}}

Multiply both sides by 1/10

{{{n=expr(1/10)(-15-ln(3)/ln(2))}}}

{{{m=3n+7}}}

{{{m=expr(3/10)(-15-ln(3)/ln(2))+7}}}

But you don't want m or n, you want m-n

{{{m-n=expr(3/10)(-15-ln(3)/ln(2))+7-expr(1/10)(-15-ln(3)/ln(2))}}}

{{{m-n=expr(2/10)(-15-ln(3)/ln(2))+7}}}

{{{m-n=expr(1/5)(-15-ln(3)/ln(2))+7}}}

Edwin</pre>