how would i Find the solution to the equation 64^(4 – x) = 4^2x

Question

e.mccormick · Answer

You want to get the bases of the powers to be the same using exponent rules.

anonymous · Answer

so in this instance can i put 2^6(4-x)= 2^2(2x)?

anonymous · Answer

and then just cancel out the bottom basses

e.mccormick · Answer

what is $4^3$?  (makes it easier)

anonymous · Answer

64

e.mccormick · Answer

So what is $(4^3)^{4 - x}$'s exponent when you are all done?

anonymous · Answer

oh, so do i get them both to be 64? and then i should have 64^(4-x)= 4^3(2x)?

anonymous · Answer

or 4^3(4-x)?=

e.mccormick · Answer

Not both to 64.  Both to a base of 4.

AH HA!  There you have it...  or what looks like it.

$4^{3(4-x)}$  so $4^{3(4-x)}= 4^{2x}$

anonymous · Answer

alright, so then the basses cancel out right, and then i have just $$3(4-x)=2x$$

anonymous · Answer

and then simplify by doing 12-3x=2x

e.mccormick · Answer

Yes!

$4^{3(4-x)}= 4^{2x}$

$\cancel{4}^{3(4-x)}= \cancel{4}^{2x}$

$3(4-x)= 2x$

e.mccormick · Answer

Yah, at that point it should be easy.

anonymous · Answer

so then 12=5x

anonymous · Answer

etc

e.mccormick · Answer

And there you have it!

anonymous · Answer

alright so in the end i have 12/5 which is 2.4
is that the correct answer?

anonymous · Answer

thank you so much for the help by the way!

e.mccormick · Answer

Yah, 12/5.  Have fun!