Question

How do you solve `\log (4x+1)-\log (x-3)=1`?

Answer 1

Use the property `log_an - log_am = log_a(n/m)`.

`log( (4x + 1)/(x - 3) )= 1`

Recall that `logx` generally signifies a logarithm in base `10`.

`(4x + 1)/(x - 3) = 10^1`

`4x + 1 = 10(x - 3)`

`4x + 1 = 10x - 30`

`31= 6x`

`x = 31/6`

Verify to make sure this is not an extraneous solution.

Hopefully this helps!