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Log (3x+5) = 1 + (log (x-1)

I need to solve this equation:   Log(3x+5) = 1 + (log (x-1)

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3 Answers

log(3x + 5) = 1 + log(x - 1)
log(3x+5) - log(x-1) = 1
log[(3x+5)/(x-1)] = log10
(3x+5)/(x-1) = 10 (x≠1)
3x + 5 = 10(x - 1)
3x - 10x = -5 - 10
-7x = -15 ---> x = 15/7

Comments

woops, my bad x = 15/7

Comments

I think there is an error here.  Since the bases are the same when you move the log(x-1) to the left, we could use the quotient rule of logs to get Log(3x+5)/(x-1)=1...which we can rewrite in exponential form; 10^1=(3x+5)/(x-1)...which gives us 10x-10=3x+5....7x=15....x=15/7 

Hi Thomas. There is no any evidence, that given logarithm is natural (ln).

Since the bases are the same when you move the log(x-1) to the left, we could use the quotient rule of logs to get Log(3x+5)/(x-1)=1...which we can rewrite in exponential form; 10^1=(3x+5)/(x-1)...which gives us 10x-10=3x+5....7x=15....x=15/7