Algebra II Help

posted by .

How do you solve this ;

5^(3y) = 8^ (y-1)

I think the first step is (3y)log 5 = (y-1)log 5

2nd step:3ylog5 = ylog5-1log5
3rd step: 3ylog5 - ylog5 = -1log5
4th step: y(2ylog5 - log5) = -1log5


From here I don't know, normal if I got rid of the 2y I would divide -1log 5 by log5-log5 ...

  • Algebra II Help -

    Solved it

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Algebra help

    f(x)=x^3-4x^2-x+4 How do I find the x-intercept for this?
  2. Algebra 2

    how do i solve these two problems? 3log(base5)^(x^2+9)-6 = 0 log(base8)^(n-3) + log(base8)^n+4)= 1 can you tell me step by step as well as the answers?
  3. Algebra 2

    Can you please explain step by step on how to do this problem?
  4. Algebra 2

    Can you please explain step by step on how to do this problem?
  5. pre algebra

    solve the equation step by step.list each step along with the property you used during that step. -5 x 1 x 11 x 4
  6. Algebra

    Please show me step by step on solving this problem?
  7. Algebra

    will you please show me step by step in solving: 1/2 log b + 3 log c. I need to express as a single log logarithm and if possible, simplify Thanks
  8. Precalculus

    This is for solving exponential/logarithmic functions: (This is a base e Logarithmic function I would assume): e^(4x)/10 =4^x-2 ?
  9. ALGEBRA HELP

    When solving the equation, what property was used to go from step 3 to step 4?
  10. Quick math help

    In which step below does a mistake first appear in simplifying the expression?

More Similar Questions