Question

`log_3(x+12)+log_3(x-12)=4` ?

Answer 1

`x=15`

Explanation:

We have:

`log_3(x+12)+log_3(x-12)=4`

We can combine the left side by saying:

`log_3((x+12)(x-12))=4`

Know that we can rewrite log questions using the following format:

`a^b=c <=> log_(a)c=b`

So we can write out question as:

`3^4=(x+12)(x-12)`

And let's expand:

`81=x^2-144`

`x^2=225`

`x=+-15`

And now let's test the solutions:

For `x=15`

`log_3(x+12)+log_3(x-12)=4`

`log_3(15+12)+log_3(15-12)=4`

`log_3(27)+log_3(3)=4`

`3+1=4`

For `x=-15`

`log_3(-15+12)+log_3(-15-12)=4`

`log_3(-3)+log_3(-27)=4` and this doesn't work because we can't have negatives inside the log function.

Therefore our solution is `x=15`

Answer 2

`x = 15`

Explanation:

I have assumed that you meant `log_3` as I have now shown in the question.

First problem:
You can't mix log terms and number terms, all one or all the other.

Second problem:
How do you write 4 as `log_3`?

Compare:
`10^1 = 10 hArr log_10 10 = 1" and "3^1 =3 hArr log_3 3=`
`10^2 = 100 hArr log_100 = 2" and " 3^2 = 9 hArr log_3 9 = 2`
`10^3 = 1000 hArr log_1000 = 3" and "3^3 = 27 hArr log_3 27 =3`

`log_3 (x+12) + log_3 (x-12) = color(red)(4)" "larr` can be written as:
`log_3 (x+12) + log_3 (x-12) = color(red)(log_3 81)`

Condense the two log terms into one:

`log_3 (x+12)(x-12) = log_3 81`

Now use the log law: If `log A = log B hArr A = B`

`cancel(log_3) (x+12)(x-12) = cancel(log_3) 81`

`(x+12)(x-12) = 81" "larr` notice the difference of squares

`x^2 -144 = 81`

`x^2 = 225`

`x = +-sqrt225 = +-15`

However in this question, `-15` is not valid.

`x = 15`

Be sure to notice the base is 3, not 10!