SOLUTION: Solve Equation log5(3x-1)=1 I can't figure this one out, any help would be appreciated!! log x+log(x-3)=1 I solved this one but wanted to make sure I'm doing this corre

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve Equation log5(3x-1)=1 I can't figure this one out, any help would be appreciated!! log x+log(x-3)=1 I solved this one but wanted to make sure I'm doing this corre      Log On


   



Question 276286: Solve Equation
log5(3x-1)=1 I can't figure this one out, any help would be appreciated!!


log x+log(x-3)=1 I solved this one but wanted to make sure I'm doing this correct

log(x(x+3))=1
x(x+3)=10^1=10
x^2+3x=10
x^2+3x-10=0
(x+5)(x-2)=0
x=-5, 2

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I'm not really sure why you would have trouble with the first equation and not the second!? The first one is easier.

In the solution of the second equation you reached an equation:
log(x(x+3))=1
and then rewrote it in exponential form:
x(x+3)=10^1=10

The first equation is already in the same form (i.e. log(expression)=other-expression):
log%285%2C+%283x-1%29%29+=+1
So we do the same thing. We rewrite it in exponential form:
3x - 1 = 5^1 = 5
And from this point on, this equation is easier to solve than the second one, too. I'll leave the rest up to you.