Here's the question you clicked on:
vivalakoda
Find the solution to the equation 64^(4 – x) = 4^(2x)?
Notice:\[64 ^{4 - x} = (4^3)^{4 - x}\]
Oh, 4^3 is equal to 64. So would I make it: \[64^{4-x}=64^{4-x}\]
:\ So look in your text book yet?
Sadly I don't have a textbook, I do online school. I've been trying to reach my teacher all morning but she hasn't responded :/
I do online school to I know how it feels when a teacher doesn't reach you back =-=
do what @ParthKohli said rewrite \(64\) as \(4^3\) this means \(64^{4-x}=4^{3(4-x)}=4^{12-3x}\)
then you know \(12-3x=2x\) so you can solve for \(x\)
do what satellite73 tells you to do if you can create the same bases then you can set the exponents equal to each other and solve for x
Thank you everyone for the help, I understand the problem now.