Question

How do you solve `2log_5x – log_5 2 = log_5 (2x+6)`?

Answer 1

`x=6`; refer below for explanation.

Explanation:

We'll have to use a few properties of logs here, namely:
(i)`..........log_bx-log_by=log_b(x/y)`
(ii)`..........alog_bx=log_b(x^a)`
Plus, if `log_bx=log_by`, then `x=y` (basically, if the bases are the same, then what's inside the logarithms on each side of the equation are equal to each other).

Luckily, all of are logarithms are base 5, so we can apply all the above properties. Using property ii on `2log_5x`, we have:
`2log_5x=log_5(x^2)`
Our equation is now `log_5(x^2)-log_5 2=log_5(2x+6)`

Using property i on this equation, this simplifies to:
`log_5(x^2/2)=log_5(2x+6)`

Because both logs have the same bases, we can say:
`x^2/2=2x+6`

This is a quadratic equation which is easily solved:
`x^2=4x+12`
`x^2-4x-12=0`
`(x-6)(x+2)=0`
`x=6` and `x=-2`

Two solutions...now what? Well, logarithms are defined only for `x>0`, and one of our solutions `(x=-2)` is less than zero. We simply get rid of it. Therefore, the solution to the equation is `x=6`.