Question 896967

Solve 16^(x+3)=64^(2x-5)
<pre>
{{{16^(x + 3) = 64^(2x - 5)}}}
{{{(4^2)^(x + 3) = (4^3)^(2x - 5)}}}
{{{4^(2(x + 3)) = 4^(3(2x - 5))}}}
2(x + 3) = 3(2x - 5) ------ Bases are equal, and so are their exponents
2x + 6 = 6x - 15
2x - 6x = - 15 - 6
- 4x = - 21
{{{x = (- 21)/(- 4)}}}, or {{{highlight_green(highlight_green(5.25))}}}
You can do the check!! 

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