Solve log &#x2061;<!-- ⁡ --> ( 2 x + 1

Davon Irwin

Davon Irwin

Answered question

2022-06-24

Solve log ( 2 x + 1 ) log 4 1 log ( 3 x + 2 ) = 1
My attempt:
log ( 2 x + 1 ) log 4 1 log ( 3 x + 2 ) = 1
log ( 2 x + 1 ) log 4 log 10 log ( 3 x + 2 ) = log 10
log ( 2 x + 1 ) 4 log 10 ( 3 x + 2 ) = log 10
( 2 x + 1 ) 4 10 ( 3 x + 2 ) = 10
( 2 x + 1 ) 4 = 10 10 ( 3 x + 2 )
(is this even correct ??)
( 2 x + 1 ) 4 = 100 ( 3 x + 2 )
( 2 x + 1 ) 100 = 4 ( 3 x + 2 )
200 x + 100 = 12 x + 8
188 x = 92
x = 23 / 47
The solution is 2.

Answer & Explanation

hopeloothab9m

hopeloothab9m

Beginner2022-06-25Added 25 answers

log ( 2 x + 1 ) log 4 1 log ( 3 x + 2 ) = 1
becomes
log ( 2 x + 1 ) log 4 = 1 log ( 3 x + 2 )
that is
log 2 x + 1 4 = log 10 3 x + 2
Can you go on from here? (I assume decimal logarithms.)
You can't go from
log a log b = log c
to
a b = c
Try some values to convince you about this.
Izabella Ponce

Izabella Ponce

Beginner2022-06-26Added 4 answers

log ( 2 x + 1 ) log 4 1 log ( 3 x + 2 ) = 1
log ( 2 x + 1 ) / 4 log 10 / ( 3 x + 2 ) = 1
log ( 2 x + 1 ) / 4 = log 10 / ( 3 x + 2 )
( 2 x + 1 ) / 4 = 10 / ( 3 x + 2 )
( 2 x + 1 ) ( 3 x + 2 ) = 40
6 x 2 + 7 x 38 = 0
x 1 , 2 = 7 ± 31 12 : x 1 = 2 , x 2 = 19 6
in real field only x 1 = 2 is a solution

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?