Question

What is the value of `x` if `log_6 48 = log_6(x + 7) + log_6(x - 1)`?

Answer 1

`{5}`

Explanation:

Combine the logarithms.

`log_6 48 = log_6 ((x + 7)(x - 1))`

If `log_an = log_am`, then `n = m`.

`48 = (x + 7)(x - 1)`

`48 = x^2 + 7x - x - 7`

`48 = x^2 + 6x- 7`

`0 = x^2 + 6x - 55`

`0 = (x+ 11)(x -5)`

`x = -11 and 5`

`x = -11` is clearly extraneous as a logarithm can never be negative. `x= 5` is valid, though.

Practice Exercises

1. Solve the following equations using `log_a n - log_a m = log_a(n/m)` and `log_a m + log_a n = log_a (m * n)`.

a) `log_2 (2x) + log_2 (1/2x) = log_2 64`
b) `log_3 (x + 2) - log_3 (x - 4) = 1`

Answers:

a) `{8}`
b) `{7}`

Hopefully this helps, and good luck!