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My homework probem is 4^(3x-3)=32. I know that I can put 4 as 2^2 and multiply the 2 (exponent) to (3x-3) but then what do I do with the 32?
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\[4^{3x-3}=32\] that is what the problem looks like
you see that 4 \[4=2^2\] so \[4^{3x-3}=2^{6x-6}\] and also \[32=2^5\] so you get \[2^{6x-6}=2^5\] and then solve \[6x-6=5\]
ohhh so both of the twos are cancelled then?
it is not "canceled" it is a fact that \[b^x=b^y\iff x=y\]
thank you! I really appreciate it!
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