<s>[INST] Solve given equation $4^{(x-2)(x+3)} - 64^{(x-3)} = 0?$ 
Solve given equation $4^{(x-2)(x+3)} - 64^{(x-3)} = 0?$


My attempt:
I've attempted to solve this question, but isn't it impossible to solve, i.e has already been simplified completely?  
$4^{(x-2)(x+3)} - 64^{(x-3)} = 0$ 
Because $(x-2)(x+3) = x^2 +x - 6?$
So adding those two powers together would mean that they are unable to be solved for $x$?
Thanks
 [/INST] From
$$
4^{x^2+x-6}-64^{x-3}=0
$$ you get
$$
2^{2(x^2+x-6)}-2^{6(x-3)}=0
$$ or
$$
2^{2(x^2+x-6)}=2^{6(x-3)}
$$ equivalently

$$
2(x^2+x-6)-6(x-3)=0
$$ 

Can you take it from here?
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