anonymous
  • anonymous
Find the solution to the equation 64^(4 – x) = 4^(2x)?
Mathematics
chestercat
  • chestercat
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ParthKohli
  • ParthKohli
Notice:\[64 ^{4 - x} = (4^3)^{4 - x}\]
anonymous
  • anonymous
Oh, 4^3 is equal to 64. So would I make it: \[64^{4-x}=64^{4-x}\]
lacypennelll
  • lacypennelll
:\ So look in your text book yet?

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anonymous
  • anonymous
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 :/
lacypennelll
  • lacypennelll
I do online school to I know how it feels when a teacher doesn't reach you back =-=
anonymous
  • anonymous
do what @ParthKohli said rewrite \(64\) as \(4^3\) this means \(64^{4-x}=4^{3(4-x)}=4^{12-3x}\)
anonymous
  • anonymous
then you know \(12-3x=2x\) so you can solve for \(x\)
precal
  • precal
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
anonymous
  • anonymous
Thank you everyone for the help, I understand the problem now.
precal
  • precal
yw

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