# Math Help - solve the logarithm equation

1. ## solve the logarithm equation

im not to sure how far i should be going to solve the equation, am i supposed to find out what x is. if so this is what i have come up with
Question: 5^x-1=2^3x-1 give answer to 3s.f

Workings
log5 ^ 23x-1=x-1
3x-1log5^2=x-1
log5^2=x-1/3x-1
1.4=x-1/3x-1
1.4(3x-1)=x-1
4.2x-1.4=x-1
3.2x-1.4=-1
3.2x=0.4
x=0.4/3.2
x=0.125
3.s.f= 0.13
i know somewhere i have gone really wrong, but not quite sure where. any tips?

2. 5^x-1=2^3x-1

(x-1)ln 5 = (3x-1) ln 3

= 1.60943791x - 1.60943791 = 3.29583687x- 1.09861229

1.68639895x = - 0.510825624

x = -0.317393806

3. (x-1)ln 5 = (3x-1) ln 3
where you have done this part here
why would it be (3x-1)ln3
and not (3x-1)ln 2
i may have missed something but would you be kind enough to give a short explanation
thankyou

4. Oh, I'm sorry. I misread that.

Follow the steps I did except use ln 2 instead of ln 3

5. Originally Posted by satx
5^x-1=2^3x-1

(x-1)ln 5 = (3x-1) ln 3

= 1.60943791x - 1.60943791 = 3.29583687x- 1.09861229

1.68639895x = - 0.510825624

x = -0.317393806
Unless your calculator is very old it's better to only get a decimal approximation at the end.

$(x-1) \ln (5) = (3x-1) \ln 3$

$x \ln (5) - \ln(5) = 3x \ln (3) - \ln (3)$

$x[\ln (5) - 3 \ln (3)] = \ln (5) - \ln (3)$

$x = \frac{\ln (5) - \ln (3)}{\ln (5) - 3 \ln (3)}$

From what I can guess it's the same answer, just that one is exact