Question 1079943
Please help not sure how to solve this equation.
(16^(x+2))/(64)=((1)/(64))^(3x-2)
<pre>{{{16^(x + 2)/64 = 1/64^(3x - 2)}}}
{{{(4^2)^(x + 2)/4^3 = 1/(4^3)^(3x - 2)}}} ------- Converting all bases to base 4
{{{4^(2x + 4 - 3) = 1/4^(9x - 6)}}} 
{{{4^(2x + 1) = 4^(- (9x - 6))}}}
2x + 1 = - 9x + 6 ------ Bases are equal and so are the exponents
2x + 9x = 6 - 1
11x = 5
{{{highlight_green(matrix(1,3, x, "=", 5/11))}}}
It'as SIMPLE as that!!
There's ABSOLUTELY NO reason, at all, to use logs!! Doesn't make sense to me why someone would!