Log Questions: Answers to x=2 & x=10^3

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This discussion addresses two logarithmic equations involving base 3 and base x. In Problem 1, the equation simplifies to show that x equals 2 after correcting the parentheses in the logarithmic expression. In Problem 2, the equation simplifies to 3 = logx, leading to the conclusion that x equals 10^3. The key steps involve applying logarithmic properties and careful manipulation of the equations.

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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!
 
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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!

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
 
Thanks buddy.
 

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