Question

How do you solve `log3x=log2+log (x+5)`?

Answer 1

`x = 10`

Explanation:

Rearrange slightly to get `x` on one side and constant on the other

`log3x = log 2 + log(x + 5)`
`log3 + logx = log 2 + log(x+5)`
`logx - log(x+5) = log2 - log3`

Now, using laws of logarithms, make it so you only have a single logarithm on either side

`logx - log(x+5) = log(x/(x+5))`
`log2 - log3 = log(2/3)`

So you have

`log(x/(x+5)) = log(2/3)`

Raise both sides to the power `e` to cancel out the logarithms (assuming we are dealing with `log_e` or `ln`),

`x/(x+5) = 2/3`

From which we get

`x/(x+5) = 2/3`
`3x = 2(x+5)`
`3x = 2x + 10`
`3x - 2x = 10`
`x = 10`