# solve for x: `2log_b x= log_b 4 + log_b(x-1).`

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### 2 Answers

your question is ,Apply laws of log and solve the problem.I hope you know laws of log.Log defined only for positive number more than 0.

`x>0 ,x-1>0` this implies `x>1` .

`clog_ba=log_ba^c,log_ba-log_bc=log_b(a/c)`

Now take your question.

`2log_bx=log_b4+log_b(x-1)`

`log_bx^2=log_b4+log_b(x-1)`

`log_bx^2-log_b(x-1)=log_b4`

`log_b(x^2/(x-1))=log_b4`

`` taking antilog both side

`x^2/(x-1)=4`

`x^2=4x-4`

`x^2-4x+4=0`

`(x-2)^2=0`

`x-2=0`

`x=2`

`x=2>1`

So answer to your question x=2

`2log_bx=log_b4+log_b(x-1)` (given)

Since,` alog_bx=log_bx^a` and ` log_bx+log_by=log_bxy` we get:

`log_bx^2=log_b4*(x-1)`

`rArr x^2=4*(x-1)`

`rArr x^2-4x+4=0`

`rArr (x-2)^2=0`

`rArr x-2=0`

`rArr x=2`

**Therefore, the required value of x is 2.**

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