Question

How do you solve `\log _ { 4} ( m - 3) + \log _ { 4} ( m + 3) = 2`?

Answer 1

`m=5`

Explanation:

`"using the "color(blue)"laws of logarithms"`

`•color(white)(x)logx+logyhArrlog(xy);x,y>0`

`•color(white)(x)log_b x=nhArrx=b^n`

`rArrlog_4(x-3)(x+3)=2`

`rArrx^2-9=4^2=16`

`rArrx^2=16+9=25`

`rArrx=+-5`

`"but "m-3>0" and "m+3>0`

`rArrx=5" is the solution ",x=-5" is invalid"`

Answer 2

`m=5`

Explanation:

All the terms need to be in the same form. Either all logs, or all numbers.

Change `2` into a log of base `4`

`log_4 x = 2 hArr x = 4^2 = 16`

`log_4 (m-3) +log_4 (m+3) = log_4 16`

Recall a law of logs: `color(blue)(log (axxb) = log a+ log b)`

Multiplying numbers is the same as adding their logs.

Adding logs is the same as multiplying the numbers.

`log_4 ((m-3)(m+3)) = log_4 16`

Recall another log law. If `color(blue)( log a = log b, " then " a = b)`

`(m-3)(m+3) = 16`

`m^2 - 9 =16`

`m^2 = 25`

`m = +-sqrt25`

`m = +5 or m = -5`

However, logs are undefined for negative values, so `-5` is not a valid solution.

`m =5`