Step by step
The question
Question: solve log₃(x) + log₃(x − 6) = 3.
One rule first: what is inside a log must be positive.
There is no log of 0 or of a negative number.
We use it in the check.
1 Make one log
The product rule, from the log rules lesson: plus outside the logs is times inside one log.
So we get one log: log₃(x(x − 6)).
Careful!
Plus between two logs is not plus inside.
It is times inside one log.
2 Write it as a power
A log is a power.
So 3 to the power 3 is what is inside.
So x(x − 6) is 3³.
3³ is 27.
3 Open the brackets
x times each part, then like terms: x² − 6x = 27.
4 Get 0 on one side
To solve it, we want 0 on the right side.
So we take away 27 on both sides.
We do it on both sides, so they stay equal.
Now we have x² − 6x − 27 = 0.
5 Factor
We need two numbers.
They multiply to −27, and they add to −6.
They multiply to a minus number, so one number is plus and one is minus.
They add to a minus number, so the bigger one is minus.
We list the pairs that make 27, smallest first.
We take the smaller from the bigger.
We need 6.
3 and 9 give 6.
Yes!
With the signs: 3 and −9.
So it is (x + 3)(x − 9) = 0.
6 One bracket is 0
Two numbers multiply to 0.
That only happens if one of them is 0.
So one bracket must be 0.
We make each bracket 0.
Two small equations.
In x + 3 = 0, we take away 3 on both sides.
In x − 9 = 0, we add 9 on both sides.
So x = −3 or x = 9.
7 Check
An answer can be false here: a log of a negative number does not exist.
So we check every answer in the question.
Let's check x = −3.
Inside a log we get −3.
There is no log of a negative number.
So x = −3 is a false answer.
Books call it extraneous.
Now x = 9.
The insides are 9 and 3: both positive.
Times them inside one log: log base 3 of 27 is 3.
It works!
Our answers were:
We cross out the false answer.
Only x = 9 is left.
Let's look back.
First: make one log.
Then: write it as a power.
Then: open the brackets.
Then: get 0 on one side.
Then: factor.
Then: one bracket is 0.
Last: check.