I have the problem 27^4z=9^(z+1) I don't want the answer, just help figuring out the proper way to simplify both sides of the equation so I can add the exponents. I know the bases will be the same, 3^(whatever power)

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I have the problem 27^4z=9^(z+1) I don't want the answer, just help figuring out the proper way to simplify both sides of the equation so I can add the exponents. I know the bases will be the same, 3^(whatever power)

Mathematics
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I have gotten as far as 3^3(4z)=3^2(z+1)
log both sides
I don't know what to do with the exponents, whether I should add them, multiply them, divide them or whatever else. I am not familiar with the log function yet. The chapter the homework is in is on exponential functions, so trying to do it that way first.

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27^4z=9^(z+1) log(27^4z) = log(9^(z+1)) 4z (log(27)) = (z+1) (log(9)) get like terms together now
I already have the bases, I just need to know what to do with the exponents. Again, trying to do this without a calculator. I want to understand the process before I rely on a calculator for the answers
exponents an logs are hand in hand. they are inverses of each other and you need logs to work exponents with a variable in them
you can "remember" exponents if you want and try to recall what is what... but it is harder
4z (log(27)) = (z+1) (log(9)) 4z.3log (3)= (z+1)2log (3) 12z=2z+2 z=1/5
the relation is this: B^x = y x = logB(y)
3^2 = 9 2 = log3(9)
As for exponent rules: 3^(2^x) = 3^(2x). when exponents have an exponent; the exponents multiply together
THAT is what I needed :D The exponent rule.
I got it right easily that time. Just didn't know what to do with it, thank you very much d:D
:)
{I have gotten as far as 3^3(4z)=3^2(z+1)}. when you got this , why are you mess with log.. just compare exponents of 3 n get the result
easy when you can "know" the base..

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