Question

how? 64^(x – 4) = 4^2x.

Answer 1

I dunno, try taking log base 2 of both sides

Answer 2

rewrite it all as powers of 2\[(2^6)^{x-4}=(2^2)^{2x}\]now applying the laws of exponents\[2^{6x-24}=2^{4x}\]and now we can take the log base 2 and solve it :)