anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I have gotten as far as 3^3(4z)=3^2(z+1)
amistre64
  • amistre64
log both sides
anonymous
  • anonymous
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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More answers

amistre64
  • amistre64
27^4z=9^(z+1) log(27^4z) = log(9^(z+1)) 4z (log(27)) = (z+1) (log(9)) get like terms together now
anonymous
  • anonymous
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
amistre64
  • amistre64
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
amistre64
  • amistre64
you can "remember" exponents if you want and try to recall what is what... but it is harder
anonymous
  • anonymous
4z (log(27)) = (z+1) (log(9)) 4z.3log (3)= (z+1)2log (3) 12z=2z+2 z=1/5
amistre64
  • amistre64
the relation is this: B^x = y x = logB(y)
amistre64
  • amistre64
3^2 = 9 2 = log3(9)
amistre64
  • amistre64
As for exponent rules: 3^(2^x) = 3^(2x). when exponents have an exponent; the exponents multiply together
anonymous
  • anonymous
THAT is what I needed :D The exponent rule.
anonymous
  • anonymous
I got it right easily that time. Just didn't know what to do with it, thank you very much d:D
amistre64
  • amistre64
:)
anonymous
  • anonymous
{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
amistre64
  • amistre64
easy when you can "know" the base..

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