Question

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)

Answer 1

I have gotten as far as 3^3(4z)=3^2(z+1)

Answer 2

log both sides

Answer 3

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.

Answer 4

27^4z=9^(z+1)

log(27^4z) = log(9^(z+1))

4z (log(27)) = (z+1) (log(9))

get like terms together now

Answer 5

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

Answer 6

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

Answer 7

you can "remember" exponents if you want and try to recall what is what... but it is harder

Answer 8

4z (log(27)) = (z+1) (log(9))
4z.3log (3)= (z+1)2log (3)
12z=2z+2
z=1/5

Answer 9

the relation is this:

B^x = y

x = logB(y)

Answer 10

3^2 = 9

2 = log3(9)

Answer 11

As for exponent rules:

3^(2^x) = 3^(2x).

when exponents have an exponent; the exponents multiply together

Answer 12

THAT is what I needed :D The exponent rule.

Answer 13

I got it right easily that time. Just didn't know what to do with it, thank you very much d:D

Answer 14

:)

Answer 15

{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

Answer 16

easy when you can "know" the base..