Question on logs

[Death]

I need some help on two log questions. I do understand everything of the problem until the final answer, it does not make sense.

Problem 1:
log3(x-1)^2 - log3(x^2 -1) = -1 + log3(x-1)

2log3(x-1) - log3(x^2 -1) = -1 + log3(x-1)

2log3(x-1) - log3(x-1) - log3(x^2 -1) = -1

log3(x-1) - log3(x^2 -1) = -1

log3 x-1/(x-1)(x+1) = -1

log3 1/x+1 = -1

=3^-1 = 1/3

x = 2 <=== why does it equal two?

Problem 2:
logx^5 + log x = (log x)(2logx)
5logx + logx = 2(logx)^2
6logx = 2(logx)^2
3 = logx <=== why does it equal 3? How does it turn from 6logx to 3?
x = 10^3

Thank you! [:D]

[Njorl]

Originally posted by Death
I need some help on two log questions. I do understand everything of the problem until the final answer, it does not make sense.

Problem 1:
log3(x-1)^2 - log3(x^2 -1) = -1 + log3(x-1)

2log3(x-1) - log3(x^2 -1) = -1 + log3(x-1)

2log3(x-1) - log3(x-1) - log3(x^2 -1) = -1

log3(x-1) - log3(x^2 -1) = -1

log3 x-1/(x-1)(x+1) = -1

log3 1/x+1 = -1

=3^-1 = 1/3

x = 2 <=== why does it equal two?

Problem 2:
logx^5 + log x = (log x)(2logx)
5logx + logx = 2(logx)^2
6logx = 2(logx)^2
3 = logx <=== why does it equal 3? How does it turn from 6logx to 3?
x = 10^3

Thank you! [:D]

Problem 1
You got a little sloppy with parentheses

log3 x-1/(x-1)(x+1) = -1

log3 1/x+1 = -1



should be

log3 (x-1)/(x-1)(x+1) = -1

log3 1/(x+1) = -1

raise three to the power of each side yields

1/(x+1)=3^(-1)=1/3

1/(x+1)=1/3
x+1=3
x=2


Problem 2

6logx = 2(logx)^2
3 = logx <=== why does it equal 3? How does it turn from 6logx to 3?


both sides were divided by 2logx

6logx/2logx=3

2(logx)^2/2logx=logx

Njorl

[Death]

Thanks buddy. [:D]