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

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**

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

Tutors, please sign in to answer this question.

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**

woops, my bad x = 15/7

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).

Sean B.

Talented Math and SAT tutor for all

New York, NY

5.0
(18 ratings)

John K.

Ardent Mathematics, Science, History, SAT and German Polymath

College Point, NY

5.0
(50 ratings)

Steven M.

Premium Test Prep and Subject Tutor - New York City UWS

New York, NY

5.0
(179 ratings)

## Comments