Question

How do you solve `log_6 x -log_6(x-6)=1`?

Answer 1

`x=7.2`

Explanation:

Using the Rule`: log_ma-log_mb=log_m(a/b)`, we get,

`log_6(x/(x-6))=1...............(1)`

Since, `log_ma=n rArr m^n=a`, we have, from `(1)`,

`x/(x-6)=6^1=6`

`:. x=6(x-6)=6x-36`

`:. 5x=36 rArr x=36/5=7.2`

This root satisfy the given eqn.

Hence, `x=7.2` is the Soln.

Answer 2

Use `log(a) - log(b) = log(a/b)` to combine to a single log, eliminate the log by making both sides a power of the base, then solve for x.

Explanation:

Use the property `log(a) - log(b) = log(a/b)`:

`log_6(x/(x - 6)) = 1`

Make the log disappear by making both sides the exponent of 6:

`x/(x - 6) = 6^1`

Solve for x:

`x = 6x - 36`

`-5x = -36`

`x = 36/5`

Check:

`log_6(36/5) - log_6(36/5 - 6) = 1`

`log_6(36) - log_6(5) - log_6(6) + log_5(5) = 1`

`2log_6(6) - log_6(5) - log_6(6) + log_5(5) = 1`

`log_6(6) = 1`

`1 = 1`

This checks.