SOLUTION: can i know how about this question: {{{ log 5^(3x-1) = log 6^(x+4) }}} I try many way but I can't balance it ><

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: can i know how about this question: {{{ log 5^(3x-1) = log 6^(x+4) }}} I try many way but I can't balance it ><      Log On


   

Question 422293: can i know how about this question:
+log+5%5E%283x-1%29+=+log+6%5E%28x%2B4%29+
I try many way but I can't balance it ><
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
+log%28%285%5E%283x-1%29%29%29+=+log%28%28+6%5E%28x%2B4%29%29%29+
We start by using a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29, to move the exponents of the arguments out in front:
(3x-1)log(5) = (x+4)log(6)
This gets the variable out of the exponents. We can now solve for x. First we simplify. Using the Distributive Property we get:
(3x)log(5) - log(5) = (x)log(6) + 4log(6)
Next we gather the x terms on one side and the other terms on the other side of the equation. Subtracting (x)log(6) and adding log(5) we get:
(3x)log(5) (x)log(6) = 4log(6) + log(5)
Factoring out x on the left side we get:
x(3log(5) - log(6)) = 4log(6) + log(5)
And then we divide by (3log(5) - log(6)):

This is the solution to your equation.

P.S. If your equation was
log%285%2C+%283x-1%29%29+=+log%286%2C+%28x%2B4%29%29
instead of
+log%28%285%5E%283x-1%29%29%29+=+log%28%28+6%5E%28x%2B4%29%29%29+
then...
  • The 5 and the 6 are the bases of the logarithms, not bases with exponents of 3x-1 and x=4.
  • The 3x-1 and x+4 are not exponents at all. They are the arguments of the logarithms.
  • Since you already know some of Algebra.com's formula syntax, use log(b, (a)) (with "b" being the base and "a" being the argument) to express logarithms of bases other than 10 or e. (Click on the "Show source" link just above this solution to see how I typed these logs.)