Question 319938
{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^(x^2)/4^(2x)=64) )  }}}
<pre><b>
Subtract exponents of 4 on the left:

{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^(x^2-2x)=64)   )}}}

Since 4*4*4 = 64 write 64 as {{{4^3}}}

{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^(x^2-2x)=4^3)   )}}}

equate the exponents of 4 on each side:

{{{x^2-2x=3}}}

{{{x^2-2x-3=0}}}

{{{(x-3)(x+1)=0}}}

{{{matrix(2,3,

  x-3=0,  "|", x+1=0,
   x=3,   "|", x=-1)}}}    

Two solutions 3 and -1

Checking x=3

{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^(x^2)/4^(2x)=64) )  }}}
{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^(3^2)/4^(2*3)=64) )  }}}
{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^9/4^6=64) )  }}}
{{{4^3=64}}}
{{{64=64}}}

Checking x=-1

{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^(x^2)/4^(2x)=64) )  }}}
{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^((-1)^2)/4^(2*(-1))=64) )  }}}
{{{drawing(100,130,-1,1,-1,.5, locate(-1,0, 4^(1)/4^(-2)=64) )  }}}
{{{4^(1-(-2))=64) )  }}}
{{{4^(1+2)=64}}}
{{{4^3=64}}}
{{{64=64}}}

So both solutions check.

Edwin</pre>