Question 1070042
{{{system(log(2,(x-2y)) = 3,log(2,(x+y)) = log(2,(8)))}}}
<pre><b>
For the first equation, we use the definition of logarithm 
which states:

the logarithm equation {{{log(B,A)=C}}} is equivalent to
the exponential equation {{{A=B^C}}}  

The first equation {{{log(2,(x-2y)) = 3}}} is equivalent to {{{x-2y=2^3}}}
and since 2<sup>3</sup>=8, 

{{{x-2y=8}}}

For the second equation we use the principle:

If {{{log(B,(P))=log(B,(Q))}}} then  {{{P=Q}}}

So the second equation becomes {{{x+y=8}}}

So now we have the system of equations:

{{{system(x-2y=8,x+y=8)}}}

which you can solve by substitution or elimination/addition.

Answer:  (x,y) = (8,0)

Edwin</pre></b>