Question 173597
{{{system((1/2)^(x+y)=16, log(x-y,8)=-3)}}}. Calculate the values of x and y
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Let's simplify the equations.

Simplifying the first equation:

{{{matrix(1,3,(1/2)^(x+y),"=",16)}}}

Write {{{(1/2)}}} as {{{(2^(-1))}}}
And write {{{16}}} as {{{2^4}}}

{{{matrix(1,3,(2^(-1))^(x+y),"=",2^4)}}}

Remove the parentheses by multiplying the
inner exponent {{{-1}}} by the outer
exponent {{{x+y}}}, getting {{{-x-y}}}

{{{matrix(1,3,2^(-x-y),"=",2^4)}}}

The base on each side is the same
positive number, which is not 1, so
we may equate the exponents:

{{{matrix(1,3, -x-y, "=", 4)}}}

Let's multiply every term by {{{-1}}}

{{{matrix(1,3, x+y, "=",-4)}}}

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Now we simplify the second equation:

{{{log(x-y,8)=-3}}}

Use the rule of logs:  {{{log(B,C)=A}}} can be rewritten {{{B^A=C}}}

{{{matrix(1,3,(x-y)^(-3), "=", 8)}}}

Write {{{8}}} as {{{2^3}}}

{{{matrix(1,3,(x-y)^(-3), "=", 2^3)}}}

Enclose both sides in parentheses and raise to the
{{{1/3}}} power:

{{{matrix(1,3,((x-y)^(-3))^(1/3), "=", (2^3)^(1/3))}}}

Multiply inner exponents by outer exponents:

{{{matrix(1,3,(x-y)^(-1), "=", 2^1)}}} 

Write the left side as {{{1/(x-y)}}} and right side as just {{{2}}} 

{{{matrix(1,3,1/(x-y), "=", 2)}}}

Multiply both sides by {{{(x-y)}}}

{{{matrix(1,3,1, "=", 2(x-y))}}}

{{{matrix(1,3,1, "=", 2x-2y)}}}

Put the right side on the left and vice-versa:

{{{matrix(1,3,2x-2y, "=", 1)}}}

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So we now have this simpler system of equations:

{{{system(x+y=-4, 2x-2y=1)}}}

I assume you know how to solve that system.
If you don't, post again asking how:

Answer:  {{{matrix(1,3,x=-7/4, ",", y=-9/4)}}}

Edwin</pre>