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Solve log_5(x) + log_5(x + 4) = 1

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

Step by step

The question

Question: solve log₅(x) + log₅(x + 4) = 1.

log5(x) + log5(x + 4) = 1

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: log₅(x(x + 4)).

log5(x(x + 4)) = 1

Careful!

Plus between two logs is not plus inside.

It is times inside one log.

Careful: not log5(x + (x + 4)) ✗

2 Write it as a power

A log is a power.

So 5 to the power 1 is what is inside.

log5(c) = k means 5k = c

So x(x + 4) is 5¹.

x(x + 4) = 51

5¹ is 5.

x(x + 4) = 5

3 Open the brackets

x times each part, then like terms: x² + 4x = 5.

x2 + 4x = 5

4 Get 0 on one side

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

So we take away 5 on both sides.

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

x2 + 4x − 5 = 5 − 5

Now we have x² + 4x − 5 = 0.

x2 + 4x − 5 = 0

5 Factor

We need two numbers.

They multiply to −5, and they add to 4.

? × ? = −5
? + ? = 4

They multiply to a minus number, so one number is plus and one is minus.

They add to a plus number, so the bigger one is plus.

× is − → one +, one −; bigger one +

We list the pairs that make 5, smallest first.

We take the smaller from the bigger.

We need 4.

1 and 5 give 4.

Yes!

With the signs: −1 and 5.

1 × 5: 5 − 1 = 4 ✓

So it is (x − 1)(x + 5) = 0.

(x − 1)(x + 5) = 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 − 1 = 0 or x + 5 = 0

In x − 1 = 0, we add 1 on both sides.

In x + 5 = 0, we take away 5 on both sides.

x − 1 + 1 = 0 + 1 or x + 5 − 5 = 0 − 5

So x = 1 or x = −5.

x = 1 or x = −5

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 = −5.

Inside a log we get −5.

There is no log of a negative number.

So x = −5 is a false answer.

Books call it extraneous.

x = −5: log5(−5) + log5(−5 + 4) → log5(−5) ✗

Now x = 1.

The insides are 1 and 5: both positive.

Times them inside one log: log base 5 of 5 is 1.

It works!

x = 1: log5(1) + log5(1 + 4)
= log5(1) + log5(5) = log5(1 × 5) = log5(5) = 1

Our answers were:

x = −5 or x = 1

We cross out the false answer.

Only x = 1 is left.

x = 1

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 = 1

Your turn

Try this one yourself, then check:

log2(x) + log2(x − 7) = 3
Show the answer
x = 8

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