Question

How do you solve `log (3x+1) = 1 + log (2x-1)`?

Answer 1

The explanation is given below.

Explanation:

`log(3x+1)=1+log(2x-1)`
`log(3x+1)=log(10)+log(2x-1)` since `log(10)=1`
`log(3x+1)=log(10(2x-1))` Product rule of logarithms
`3x+1 = 10(2x-1)` if logA = log B then A= B
`3x+1 = 20x-10` distributive law

Now we have to solve for x using inverse operaitions to isolate x.

`3x+1 + 10 = 20x-10+10`

Adding 10 on both sides would remove `10` from the right side of the equation

`3x+11=20x`
`3x+11-3x=20x-3x`

Subtracting `3x` on both sides would isolate the term containing `x` to one side of the equation.

`11=17x`

Dividing both sides by 17 we get

`11/17=(17x)/17`

`11/17 = x`

`x=11/17` Answer