Question

solve: 5^(3x-4) = 15^(2x+1)

Answer 1

Someone please explain how to do this. I know I have to use logs somehhow

Answer 2

I woud ln both sides:

ln(5^(3x-4)) = ln(15^(2x+1))

then you get:

(3x - 4) * ln(5) = (2x + 1) * ln(15)

3x - 4 = 2x* ln15/ln5 + ln15/ln5

x(3 - 2ln15/ln5) = 4 + ln 15/ln5

x = (4 + ln15/ln5)/(3 - ln 225 / ln 5)

Answer 3

IS there not a way to get both sides of the equation to have the same base and then just set the exponents equal to each other?

Answer 4

hmmm, I suppose you could do this:

5^(3x-4) = 3^(2x + 1) * 5^ (2x + 1)

Then, 3^(2x+1) =

\[5^(\log_{5} 3)(2x+1)\]

then it becomes:

5^(3x-4) = \[5^(2x +1 + (\log_{5} 3)(2x+1))\]


then you can set it equal:

3x-4 = 2x +1 + (\log_{5} 3)(2x+1)

Answer 5

Thanks :)