Solve log_3(3x+2)=log_9(4x+5) for x

Kameron Wang

Kameron Wang

Answered question

2022-11-10

Solve log 3 ( 3 x + 2 ) = log 9 ( 4 x + 5 ) for x
I changed the bases of the logs
log 10 ( 3 x + 2 ) log 10 ( 3 ) = log 10 ( 4 x + 5 ) log 10 ( 9 )
Now I'm stuck, I don't know how to eliminate the logs.
On WolframAlpha I've seen that log 10 ( 3 x + 2 ) log 10 ( 3 ) = 0 gives x = 1 3 .
Do you know how does this equation and the equation above can be solved?
EDIT: You all solved the first equation, but I still don't understand how to solve the second one (the one solved by WolframAlpha).

Answer & Explanation

klofnu7c2

klofnu7c2

Beginner2022-11-11Added 14 answers

All you need is log 9 = 1 2 log 3 :
log 3 ( 3 x + 2 ) = 1 2 log 3 ( 4 x + 5 ) ( 3 x + 2 ) 2 = 4 x + 5
Could you proceed?
Humberto Campbell

Humberto Campbell

Beginner2022-11-12Added 4 answers

log 3 ( 3 x + 2 ) = log 9 ( 4 x + 5 )
Convert the RHS to base 3 to get
log 3 ( 3 x + 2 ) = log 3 ( 4 x + 5 ) log 3 3 2
So that you get
2 log 3 ( 3 x + 2 ) = log 3 ( 4 x + 5 )
The power law for logarithms yields
log 3 ( 3 x + 2 ) 2 = log 3 ( 4 x + 5 )
Now, "cancelling out the logarithms" gives you
( 3 x + 2 ) 2 = 4 x + 5
which is an easy quadratic to solve.

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