Solving exponential equations with logarithms — Mathematics

Before this

This leads to

Modelling with exponentials · coming soon
Half-life · coming soon
The natural logarithm · coming soon

An unknown that algebra cannot reach

Here is an equation that every method you have learned so far fails to solve:

3^{x} = 17.

Try the usual tools. There is nothing to expand or factorise, and $x$ is already alone on one side. Every one of those methods treats $x$ as a quantity: something you add, multiply or square. Here $x$ is something different. It is an index. It counts how many factors of 3 are multiplied together.

That is the disruption. The unknown has moved upstairs, into the index, and ordinary algebra cannot reach it there.

Equations like this are common. Money earning interest, a growing population and a decaying radioactive sample all change by the same factor again and again. For each of them, the question "how long until…?" puts the unknown in the index. The last lecture promised a tool for exactly this job. It takes about four lines.

Two answers that must be the same

We already know one answer. A logarithm is an index, so the definition answers the equation directly:

3^{x} = 17 means exactly the same as x = lo g_{3} 17 .

That answer is exact and correct. But it is not yet a number you can use. Ask the question from two lectures ago: 3 to the power of what gives 17? We can bracket it. $3^{2} = 9$ is too small, and $3^{3} = 27$ is too big. So $x$ is between 2 and 3. The figure shows where.

The curve

y = 3^{x}

passes through

(2, 9)

and

(3, 27)

. The line

y = 17

crosses it once, between

x = 2

and

x = 3

. Its exact position is

x = lo g_{3} 17

, about 2.58. The next section finds that number with algebra instead of a picture.

The lecture on what a logarithm is promised a way to find any logarithm using only the log key. Here it is.

Bring the unknown down

The method has one key move: take the logarithm of both sides. Any base works, as you will see. We use base 10, because every scientific calculator has a $lo g_{10}$ key.

Why is this allowed? The two sides of $3^{x} = 17$ are the same number. A logarithm is a function, so the same input always gives the same output. This is the same reason you may square both sides of an equation. There is one condition. Both sides must be positive, because a logarithm of zero or of a negative number does not exist.

lo g_{10} (3^{x}) = lo g_{10} 17

Now the third law. The logarithm of a power brings the power to the front:

x lo g_{10} 3 = lo g_{10} 17

Look at what just happened. The $x$ is no longer an index. It has come down to the ground floor, and now it is just multiplying a number. And $lo g_{10} 3$ is only a number, about 0.4771. So this is a linear equation of the simplest kind: $x$ times a number equals a number.

That is it. That is the whole idea. The third law turns an index into a multiplier, and multipliers are easy to undo. Divide both sides by $lo g_{10} 3$ :

x = \frac{lo g _{10} 17}{lo g _{10} 3} \approx \frac{1.2304}{0.4771} \approx 2.579

Check it. $3^{2.579} \approx 17.00$ on a calculator. ✓ It also lies between 2 and 3, as the figure showed. Notice something surprising. 17 is nearer to 9 than to 27, but $x$ is nearer to 3 than to 2. This is because the curve gets steeper as it climbs. So a bracket only says the answer is in the right interval. The check by substitution is the one to trust.

Solving an exponential equation

To solve $a^{x} = b$ , where $a$ is positive and not 1, and $b$ is positive:

1. Make sure the power stands alone on one side.

2. Take $lo g_{10}$ of both sides: $lo g_{10} (a^{x}) = lo g_{10} b$ .

3. Use the third law: $x lo g_{10} a = lo g_{10} b$ .

4. Divide: $x = \frac{lo g _{10} b}{lo g _{10} a}$ . Then calculate, and check by substituting.

Step 1 matters more than it looks. The next box shows why.

Stop and think

A student solves $2^{x} + 5 = 37$ by taking logarithms of every term: $x lo g_{10} 2 + lo g_{10} 5 = lo g_{10} 37$ . This gives $x \approx 2.89$ . What has gone wrong?

Show the answer

Taking logarithms is something you do to a whole side, not to each term. The left side is the number $2^{x} + 5$ , so its logarithm is $lo g_{10} (2^{x} + 5)$ . The laws of logarithms cannot split the logarithm of a sum. The fix is step 1: get the power alone first. Subtract 5 to get $2^{x} = 32$ . Then $x = 5$ , exactly, because $2^{5} = 32$ . Check the student's answer: $2^{2.89} + 5 \approx 12.4$ , not 37.

What a quotient of two logarithms means

We now have two answers to the same equation. The definition says $x = lo g_{3} 17$ . The method says $x = \frac{lo g _{10} 17}{lo g _{10} 3}$ . They answer the same question, so they must be equal:

lo g_{3} 17 = \frac{lo g _{10} 17}{lo g _{10} 3} .

The last lecture's Trap 2 met a quotient of two logarithms, $\frac{lo g _{10} 8}{lo g _{10} 2}$ , and found that it equals 3. It promised to show what such a quotient means. Here is the answer. $\frac{lo g _{10} 8}{lo g _{10} 2}$ is $lo g_{2} 8$ , and $lo g_{2} 8 = 3$ because $2^{3} = 8$ .

Nothing about 3 and 17 was special, so the same argument works for any allowed base. Let $c = lo g_{a} b$ . By the definition, $a^{c} = b$ . Take $lo g_{10}$ of both sides and use the third law: $c lo g_{10} a = lo g_{10} b$ . Divide by $lo g_{10} a$ , and you have the rule.

The change of base rule

For allowed bases $a$ and $k$ (positive and not 1) and positive $b$ :

$lo g_{a} b = \frac{lo g _{k} b}{lo g _{k} a}$

The base $k$ can be anything allowed, as long as the top and the bottom use the same one.

So a calculator with only a $lo g_{10}$ key can find any logarithm. For example, $lo g_{7} 100 = \frac{2}{lo g _{10} 7} \approx 2.37$ . Most calculators also have a key marked ln. It is the logarithm to a special base, $e \approx 2.718$ , which Pre-University mathematics explains. Try it: $ln 17 \div ln 3$ also gives about 2.579. The base you choose for the conversion does not matter.

Why not? The rule with $a = 2$ and $k = 10$ says that $lo g_{2} x = \frac{1}{lo g _{10} 2} \times lo g_{10} x$ , which is about $3.32 lo g_{10} x$ . Every base-2 logarithm is the base-10 logarithm multiplied by the same number. In the language of the graph transformations lectures, the graph of $y = lo g_{2} x$ is the graph of $y = lo g_{10} x$ after a vertical stretch, scale factor about 3.32.

The graph of

y = lo g_{2} x

(teal) is the graph of

y = lo g_{10} x

(ink) stretched vertically by a scale factor of about 3.32. At

x = 8

lo g_{10} 8 \approx 0.90

and

lo g_{2} 8 = 3

. At

x = 0.5

, the stretch moves the point down, from about

- 0.30

- 1

. Both graphs cross the

x

-axis at 1, which a vertical stretch leaves fixed.

So all logarithm graphs are the same shape, stretched by different amounts. In a quotient of two logarithms, the stretch factor appears on the top and on the bottom, and it cancels.

The method at work

Worked Example 1 — Using one remembered value

Solve $5^{x} = 200$ . Give your answer to 3 significant figures (3 s.f.).

The power is already alone. Take $lo g_{10}$ of both sides and bring the $x$ down: $x lo g_{10} 5 = lo g_{10} 200$ . So $x = \frac{lo g _{10} 200}{lo g _{10} 5}$ .

The last lecture built logarithms from $lo g_{10} 2 \approx 0.30103$ . Use that again. $200 = 2 \times 100$ , so $lo g_{10} 200 = 0.30103 + 2 = 2.30103$ . And $5 = 10 \div 2$ , so $lo g_{10} 5 = 1 - 0.30103 = 0.69897$ .

$x \approx 2.30103 \div 0.69897 \approx 3.292$ , so $x \approx 3.29$ (3 s.f.).

Check: $5^{3} = 125$ and $5^{4} = 625$ , so the answer must be between 3 and 4. ✓ And $5^{3.292} \approx 200.0$ . ✓

Worked Example 2 — An index with two terms

Solve $2^{x + 1} = 50$ , to 3 s.f.

The whole index is $x + 1$ . When the third law brings it down, the whole of it comes down, so it needs brackets: $(x + 1) lo g_{10} 2 = lo g_{10} 50$ .

Divide by $lo g_{10} 2$ : $x + 1 = \frac{lo g _{10} 50}{lo g _{10} 2} \approx \frac{1.69897}{0.30103} \approx 5.644$ . So $x \approx 4.644$ , and $x \approx 4.64$ (3 s.f.).

Check another way. $2^{x + 1} = 2 \times 2^{x}$ by the first index law, so the equation says $2^{x} = 25$ . Then $x = lo g_{2} 25 \approx 4.644$ . ✓ Two routes, one answer.

Worked Example 3 — A half-life that is not a whole number of halvings

A radioactive sample has a half-life of 12 years: every 12 years, half of what remains decays. A sample starts at 800 g (invented). How long until only 30 g is left?

After $t$ years, the sample has halved $\frac{t}{12}$ times. So the mass is $800 \times (\frac{1}{2})^{t /12}$ . We want $800 \times (\frac{1}{2})^{t /12} = 30$ .

Step 1, get the power alone: $(\frac{1}{2})^{t /12} = \frac{30}{800} = 0.0375$ .

Take logarithms and bring the index down: $\frac{t}{12} lo g_{10} 0.5 = lo g_{10} 0.0375$ . So $\frac{t}{12} = \frac{lo g _{10} 0.0375}{lo g _{10} 0.5} \approx \frac{- 1.42597}{- 0.30103} \approx 4.737$ .

Both logarithms are negative, because 0.0375 and 0.5 are less than 1, so their quotient is positive. Then $t \approx 12 \times 4.737 \approx 56.8$ years.

Check. After 48 years (4 halvings) there is 50 g, and after 60 years (5 halvings) there is 25 g. 30 g is between them. ✓ Counting halvings gives only the bracket. The logarithm gives the exact time.

When the unknown is in two indices

Sometimes $x$ appears in two indices with different bases, as in $5^{x} = 2^{x + 3}$ . No single logarithm answers this, because there is no one base to ask about. But the key move still works. Take logarithms of both sides, and both indices come down. What is left is a linear equation. Step through it below, and read each reason before you go on.

Step through — solving 5ˣ = 2ˣ⁺³

Press Next step to add one line. Each line comes with the reason it is allowed.

The solution of $5^{x} = 2^{x + 3}$ , one line at a time, with the reason for each line.

$lo g_{10} (5^{x}) = lo g_{10} (2^{x + 3})$ . Both sides are positive, and equal numbers have equal logarithms.
$x lo g_{10} 5 = (x + 3) lo g_{10} 2$ . The third law on each side. The whole index $x + 3$ comes down, so it needs brackets.
$x lo g_{10} 5 = x lo g_{10} 2 + 3 lo g_{10} 2$ . Expand the bracket. Now $x$ appears only as a multiplier.
$x lo g_{10} 5 - x lo g_{10} 2 = 3 lo g_{10} 2$ . Collect the terms in $x$ on one side, as in any linear equation.
$x (lo g_{10} 5 - lo g_{10} 2) = 3 lo g_{10} 2$ . Factorise, so that $x$ appears once.
$x = \frac{3 lo g _{10} 2}{lo g _{10} 2.5}$ . Divide. The second law turns $lo g_{10} 5 - lo g_{10} 2$ into $lo g_{10} 2.5$ .
$x \approx 0.90309 \div 0.39794 \approx 2.269$ . Only now use the calculator.

So $x \approx 2.27$ (3 s.f.). Check: $5^{2.269} \approx 38.6$ and $2^{5.269} \approx 38.6$ . ✓

What should you have seen? After the second line there were no indices left, only a linear equation. The logarithm's only job was to bring the unknowns down.

How long until it doubles?

The most common use of this method is a growth question. An amount grows by the same percentage each year, such as savings with interest added once a year, or a town's population. How long until it doubles?

Suppose it grows by 7% a year. Each year it is multiplied by 1.07, so after $t$ years it has been multiplied by $1.0 7^{t}$ . Doubling means $1.0 7^{t} = 2$ , and the method gives $t = \frac{lo g _{10} 2}{lo g _{10} 1.07} \approx 10.24$ years. The starting amount does not appear at all: $500 and $5 million double in the same time.

The model below lets you change the rate and the target. It is an invented, simple model. It assumes the rate stays the same every year, which real interest rates and real populations do not.

Real-world model — how long until it doubles?

Choose a growth rate per year and a target: double, triple or ten times the start. The curve shows how many times bigger the amount is after t years. Read where it meets the target line, then compare the logarithm's answer with whole years.

Growth rate: 7% a year

Target:

Presets:

How long an amount takes to reach a target when it grows by the same percentage every year (an invented, simple model with a fixed rate).

Rate a year	Target	Years (by logarithms)
2%	× 2	≈ 35.00
7%	× 2	≈ 10.24
20%	× 2	≈ 3.80
5%	× 3	≈ 22.52
10%	× 10	≈ 24.16

At 7%, after 10 whole years the amount is about 1.97 times the start, and after 11 years about 2.10 times. The "rule of 70" estimates a doubling time of $70 \div 7 = 10$ years. It is close for small rates, and less accurate for large ones: at 20% it gives 3.5 years against about 3.80.

What should you have seen? Halving the rate roughly doubles the doubling time, but not exactly. The answer is almost never a whole number of years. If growth is added once a year, the amount first passes the target at the end of the next whole year. You may know the "rule of 70" from the economic growth lecture. This is where it comes from: for small rates, $lo g_{10} 2 \div lo g_{10} (1 + r)$ is close to $70 \div r$ , with $r$ in per cent.

When the question asks "how many whole years?"

Worked Example 4 — An inequality, and a sign that flips

A machine loses 15% of its value every year (invented). After how many whole years is it worth less than a quarter of its price when new?

Each year its value is multiplied by 0.85. So we need the smallest whole number $n$ with $0.8 5^{n} < 0.25$ .

Take logarithms of both sides. A bigger positive number has a bigger logarithm, so the inequality keeps its direction: $n lo g_{10} 0.85 < lo g_{10} 0.25$ .

This is the hard step. Spend time here. $lo g_{10} 0.85 \approx - 0.07058$ is negative, because 0.85 is less than 1. Dividing both sides of an inequality by a negative number reverses it. So $n > \frac{lo g _{10} 0.25}{lo g _{10} 0.85} \approx \frac{- 0.60206}{- 0.07058} \approx 8.53$ .

The smallest whole number above 8.53 is 9. The machine is worth less than a quarter of its price after 9 years.

Check: $0.8 5^{8} \approx 0.272$ is still more than 0.25, and $0.8 5^{9} \approx 0.232$ is less. ✓ Without the flip, you would get $n < 8.53$ , which would mean the machine is worth less than a quarter when it is new. That is clearly false.

Three classic mistakes

Trap 1 — Combining the number in front with the base

$2 \times 3^{x}$ is not $6^{x}$ . The index belongs only to the 3, so $2 \times 3^{x}$ means 2 times a power of 3. To solve $2 \times 3^{x} = 54$ , divide by 2 first: $3^{x} = 27$ , so $x = 3$ . The mistake $6^{x} = 54$ gives $x \approx 2.23$ . Check it: $2 \times 3^{2.23} \approx 23$ , not 54.

Trap 2 — Rounding the logarithms too early

In $3^{x} = 17$ , rounding first to $lo g_{10} 17 \approx 1.23$ and $lo g_{10} 3 \approx 0.48$ gives $x \approx 2.5625$ , so 2.56 to 3 s.f. The true answer is 2.58. Keep the full calculator values, and round only the final answer.

Trap 3 — Expecting a solution that does not exist

$2^{x} = - 8$ has no solution. A positive base raised to any power gives a positive number. If you take logarithms anyway, you meet $lo g_{10} (- 8)$ , which does not exist. When the power stands alone and the other side is zero or negative, say that there is no solution.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Solve, giving each answer to 3 s.f.: (a) $4^{x} = 30$ (b) $7^{x} = 0.2$ (c) $3^{x - 2} = 40$ .

Hint 1
In each, the power is already alone. Take $lo g_{10}$ of both sides.

Hint 2
In (c), the whole index $x - 2$ comes down: $(x - 2) lo g_{10} 3 = lo g_{10} 40$ .

Full answer
(a) $x = \frac{lo g _{10} 30}{lo g _{10} 4} \approx \frac{1.47712}{0.60206} \approx 2.45$ . (b) $x = \frac{lo g _{10} 0.2}{lo g _{10} 7} \approx \frac{- 0.69897}{0.84510} \approx - 0.827$ . It is negative because 0.2 is less than 1, so the index must be negative. (c) $x - 2 = \frac{lo g _{10} 40}{lo g _{10} 3} \approx 3.358$ , so $x \approx 5.36$ . Check (c): $3^{3.358} \approx 40.0$ . ✓
Get the power alone first, then solve: (a) $5 \times 2^{x} = 80$ (b) $3^{x} + 7 = 50$ (c) $6 \times 1. 5^{x} - 4 = 50$ . Give exact answers where you can, and 3 s.f. otherwise.

Hint 1
Undo the adding and multiplying around the power, in the reverse order, before any logarithm.

Hint 2
(a) gives $2^{x} = 16$ . (c) gives $6 \times 1. 5^{x} = 54$ , then $1. 5^{x} = 9$ .

Full answer
(a) $2^{x} = 16$ , so $x = 4$ exactly, because $2^{4} = 16$ . No calculator is needed. (b) $3^{x} = 43$ , so $x = \frac{lo g _{10} 43}{lo g _{10} 3} \approx 3.42$ . (c) $1. 5^{x} = 9$ , so $x = \frac{lo g _{10} 9}{lo g _{10} 1.5} \approx 5.42$ . Check (c): $6 \times 1. 5^{5.419} - 4 \approx 50.0$ . ✓
(a) Find $lo g_{3} 50$ to 3 s.f. (b) Find $lo g_{5} 0.04$ exactly. (c) Show that $lo g_{2} 9 \times lo g_{3} 2 = 2$ .

Hint 1
Use the change of base rule with base 10 for (a) and (c). For (b), write 0.04 as a fraction.

Hint 2
(b) $0.04 = \frac{1}{25} = 5^{- 2}$ . (c) Write both logarithms as quotients of base-10 logarithms, and look for cancelling.

Full answer
(a) $\frac{lo g _{10} 50}{lo g _{10} 3} \approx \frac{1.69897}{0.47712} \approx 3.56$ . Check: $3^{3} = 27$ and $3^{4} = 81$ , so it must be between 3 and 4. ✓ (b) $0.04 = 5^{- 2}$ , so $lo g_{5} 0.04 = - 2$ . (c) $lo g_{2} 9 \times lo g_{3} 2 = \frac{lo g _{10} 9}{lo g _{10} 2} \times \frac{lo g _{10} 2}{lo g _{10} 3} = \frac{lo g _{10} 9}{lo g _{10} 3} = lo g_{3} 9 = 2$ . The $lo g_{10} 2$ cancels, and $3^{2} = 9$ .
Two invented towns. Town A has 20 000 people and grows by 5% a year. Town B has 35 000 people and grows by 2% a year. Both rates stay the same. In which year does Town A first have more people than Town B?

Hint 1
After $t$ years, A has $20000 \times 1.0 5^{t}$ people and B has $35000 \times 1.0 2^{t}$ . Set them equal.

Hint 2
Divide both sides by $20000 \times 1.0 2^{t}$ . This gives $(\frac{1.05}{1.02})^{t} = 1.75$ , with the power alone.

Full answer
$20000 \times 1.0 5^{t} = 35000 \times 1.0 2^{t}$ . Dividing as in Hint 2 gives $(\frac{1.05}{1.02})^{t} = 1.75$ . Take logarithms: $t = \frac{lo g _{10} 1.75}{lo g _{10} ( 1.05 \div 1.02 )} \approx \frac{0.24304}{0.01259} \approx 19.3$ . So A overtakes B during the 20th year. Check: after 19 years A has about 50 539 and B about 50 988, so B is still bigger. After 20 years A has about 53 066 and B about 52 008. ✓ The faster-growing town always wins in the end, however small it starts. The logarithm says exactly when.

The bigger idea

Exponential equations appear wherever something changes by the same factor in each equal step of time. Radioactive material decays this way, because each atom has the same chance of decaying in each year, whatever the others do.

Here is a real example. Living things take in carbon, including a tiny share of radioactive carbon-14. After death no new carbon-14 arrives, and what is there decays with a half-life of about 5730 years. A bone with 30% of its carbon-14 left solves $(\frac{1}{2})^{t /5730} = 0.3$ , so $t = 5730 \times \frac{lo g _{10} 0.3}{lo g _{10} 0.5}$ , which is about 9950 years. Willard Libby received the 1960 Nobel Prize in Chemistry for this method. Real laboratories also correct for past changes in the air's carbon-14.

This lecture ends the algebra strand of Upper Secondary, and the five lectures of this topic form one chain. The index laws counted factors, negative and fractional indices extended them, the logarithm named the index, and its laws read the index laws backwards. Today the third law brought the unknown down. At Pre-University you will meet the base $e$ , the one base for which the rate of growth of $e^{x}$ equals $e^{x}$ itself. Its logarithm is ln, and everything in this lecture works for it unchanged.

Key concepts in this lesson

To solve an equation with the unknown in an index, get the power alone, take logarithms of both sides, and use the third law to bring the index down as a multiplier.
Taking logarithms is allowed because equal positive numbers have equal logarithms, and it is done to whole sides, never term by term.
The change of base rule, $lo g_{a} b = \frac{lo g _{k} b}{lo g _{k} a}$ , lets a calculator with one logarithm key find a logarithm in any base.
All logarithm graphs are vertical stretches of one another, so the base used for the conversion cancels and does not change the answer.
Dividing an inequality by the logarithm of a number less than 1 reverses it, because that logarithm is negative.
An equation such as $2^{x} = - 8$ has no solution, because a positive base raised to any power is positive.