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Solve log_2(x) + log_2(x - 6) = 4

Step by step, with the reason for every step · Log equations with false answers · Algebra 2

Step by step

The question

Question: solve log₂(x) + log₂(x − 6) = 4.

log2(x) + log2(x − 6) = 4

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.

inside a log: > 0

1 Make one log

The product rule, from the log rules lesson: plus outside the logs is times inside one log.

log a + log b = log(a × b)

So we get one log: log₂(x(x − 6)).

log2(x(x − 6)) = 4

Careful!

Plus between two logs is not plus inside.

It is times inside one log.

Careful: not log2(x + (x − 6)) ✗

2 Write it as a power

A log is a power.

So 2 to the power 4 is what is inside.

log2(c) = k means 2k = c

So x(x − 6) is 2⁴.

x(x − 6) = 24

2⁴ is 16.

x(x − 6) = 16

3 Open the brackets

x times each part, then like terms: x² − 6x = 16.

x2 − 6x = 16

4 Get 0 on one side

To solve it, we want 0 on the right side.

So we take away 16 on both sides.

We do it on both sides, so they stay equal.

x2 − 6x − 16 = 16 − 16

Now we have x² − 6x − 16 = 0.

x2 − 6x − 16 = 0

5 Factor

We need two numbers.

They multiply to −16, and they add to −6.

? × ? = −16
? + ? = −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.

× is − → one +, one −; bigger one −

We list the pairs that make 16, smallest first.

We take the smaller from the bigger.

We need 6.

2 and 8 give 6.

Yes!

With the signs: 2 and −8.

1 × 16: 16 − 1 = 15
2 × 8: 8 − 2 = 6 ✓

So it is (x + 2)(x − 8) = 0.

(x + 2)(x − 8) = 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.

A × B = 0 → A = 0 or B = 0

We make each bracket 0.

Two small equations.

x + 2 = 0 or x − 8 = 0

In x + 2 = 0, we take away 2 on both sides.

In x − 8 = 0, we add 8 on both sides.

x + 2 − 2 = 0 − 2 or x − 8 + 8 = 0 + 8

So x = −2 or x = 8.

x = −2 or x = 8

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.

check every answer

Let's check x = −2.

Inside a log we get −2.

There is no log of a negative number.

So x = −2 is a false answer.

Books call it extraneous.

x = −2: log2(−2) + log2(−2 − 6) → log2(−2) ✗

Now x = 8.

The insides are 8 and 2: both positive.

Times them inside one log: log base 2 of 16 is 4.

It works!

x = 8: log2(8) + log2(8 − 6)
= log2(8) + log2(2) = log2(8 × 2) = log2(16)
= 4

Our answers were:

x = −2 or x = 8

We cross out the false answer.

Only x = 8 is left.

x = 8

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.

Answer
x = 8

Your turn

Try this one yourself, then check:

log3(x) + log3(x − 8) = 2
Show the answer
x = 9

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