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A. Exponential and Logarithmic Functions · Exponential and Logarithmic Functions

Solving Exponential Equations with Logarithms

A3.2 A3.1 A3.4About 75 minutes
Learning goals
  • I can solve an exponential equation by isolating the power and taking the logarithm of both sides.
  • I can solve equations with different bases on the two sides, such as 2ˣ = 3^(x − 1).
  • I can give exact answers (such as x = log 50 ÷ log 3) and decimal approximations.
  • I can solve exponential equations of quadratic type by substitution, rejecting impossible cases.

Words to know

take the logarithm of both sides
Apply the same log (usually log or ln) to both sides of an equation. This is allowed because equal positive numbers have equal logarithms.
exact answer
An answer written with logs, fractions or radicals and no rounding, such as x = log 7 ÷ log 2.
quadratic type
An equation such as 4ˣ − 6(2ˣ) + 8 = 0 that becomes a quadratic after a substitution like u = 2ˣ.
inadmissible root
A root of the substituted equation that cannot happen in the original, such as 2ˣ = −3.

When there is no common base

Try to write 3ˣ = 50 with a common base. 50 is not a power of 3, so the method of lesson 11 fails. In lesson 3 you solved equations like this by rewriting them in logarithmic form: x = log₃ 50. In this lesson you will learn a method that works for every exponential equation, even ones like 2ˣ = 3^(x − 1) with a different base on each side: take the logarithm of both sides and use the power law to bring the variable down out of the exponent.

Can both sides be writtenwith the same base?Rewrite with that base,then equate exponents1. Isolate the power2. Take log of both sides3. Power law: bring exponent down4. Solve the linear equation5. Check (exact, then decimal)yesno
A decision guide for exponential equations. Look for a common base first; if there isn't one, isolate the power and take the logarithm of both sides.
Worked example

Example 1: Take the log of both sides

Solve 3ˣ = 50. Give the exact answer and the answer to three decimal places.

  1. Take the common log of both sides: log (3ˣ) = log 50.
  2. Power law: x log 3 = log 50.
  3. Divide: x = log 50 ÷ log 3.
  4. x ≈ 1.69897 ÷ 0.47712 ≈ 3.561.
  5. Check: 3^3.561 ≈ 50.0 ✓

x = log 50 ÷ log 3 ≈ 3.561. (This is the same as log₃ 50 from lesson 3: the change of base formula and this method agree.)

Worked example

Example 2: A linear exponent

Solve 5^(2x − 1) = 70.

  1. log (5^(2x − 1)) = log 70.
  2. Power law (keep the brackets!): (2x − 1) log 5 = log 70.
  3. 2x − 1 = log 70 ÷ log 5 ≈ 2.6397.
  4. 2x ≈ 3.6397, so x ≈ 1.820.

x = (log 70 ÷ log 5 + 1) ÷ 2 ≈ 1.820

Worked example

Example 3: Different bases on each side

Solve 2ˣ = 3^(x − 1).

  1. Take logs: x log 2 = (x − 1) log 3.
  2. Expand: x log 2 = x log 3 − log 3.
  3. Collect the x-terms: log 3 = x log 3 − x log 2 = x(log 3 − log 2).
  4. x = log 3 ÷ (log 3 − log 2) ≈ 0.47712 ÷ 0.17609 ≈ 2.710.
  5. Check: 2^2.710 ≈ 6.54 and 3^1.710 ≈ 6.54 ✓

x = log 3 ÷ (log 3 − log 2) ≈ 2.710. The exact answer can also be written log 3 ÷ log 1.5.

xy−1123424681012(2.71, 6.54)y = 3^(x − 1)y = 2ˣx ≈ 2.71
The graphs of y = 2ˣ and y = 3^(x − 1) cross where x ≈ 2.71, the solution of Example 3. For larger x the base-3 curve pulls ahead.

Isolate before you log

In 4(1.06)ᵗ = 10, divide by 4 before taking logs. You could take logs right away (log 4 + t log 1.06 = log 10), but that adds a step. Never write log (4(1.06)ᵗ) = t log (4 × 1.06): the power belongs only to 1.06.

Worked example

Example 4: Isolate the power first

An investment of $4000 grows at 6% per year compounded annually, so A = 4000(1.06)ᵗ. Solve 4000(1.06)ᵗ = 10 000 to find when it reaches $10 000.

  1. Divide by 4000: 1.06ᵗ = 2.5.
  2. Take logs: t log 1.06 = log 2.5.
  3. t = log 2.5 ÷ log 1.06 ≈ 15.73.

t ≈ 15.7 years. Because interest is added once a year, the balance first passes $10 000 at the end of year 16.

Worked example

Example 5: Quadratic type

Solve (a) 4ˣ − 6(2ˣ) + 8 = 0 and (b) 9ˣ − 4(3ˣ) − 5 = 0.

  1. (a) Since 4ˣ = (2ˣ)², let u = 2ˣ: u² − 6u + 8 = 0, so (u − 2)(u − 4) = 0.
  2. (a) 2ˣ = 2 gives x = 1; 2ˣ = 4 gives x = 2.
  3. (b) Let u = 3ˣ: u² − 4u − 5 = 0, so (u − 5)(u + 1) = 0.
  4. (b) 3ˣ = −1 is impossible, because 3ˣ > 0 for every real x. Reject it.
  5. (b) 3ˣ = 5 gives x = log 5 ÷ log 3 ≈ 1.465.

(a) x = 1 or x = 2 (b) x = log 5 ÷ log 3 ≈ 1.465

Just for fun

The Royal Canadian Mint stopped making pennies in 2012. That has not stopped teachers from using doubling pennies in exponential lessons, as you are about to see.

Hands-on activity

The doubling-penny deal

You need: a calculator, paper, and (optional) a handful of coins.

A relative offers you a choice for a 30-day summer job: (A) $1000 every day, or (B) 1 cent on day 1, 2 cents on day 2, 4 cents on day 3, doubling each day. On day n, offer B pays 0.01 × 2^(n − 1) dollars.

  1. Use coins to model the first five days of offer B (1, 2, 4, 8, 16 cents). Which offer looks better so far?
  2. Write and solve an equation with logarithms to find the first day on which B's daily payment is more than $1000.
  3. With your calculator, find B's payment on day 30.
  4. Which offer would you choose?

What to look for: 0.01 × 2^(n − 1) > 1000 means 2^(n − 1) > 100 000, so n − 1 > log 100 000 ÷ log 2 ≈ 16.61. The first whole-number day is day 18 (payment $1310.72). On day 30 the payment is 0.01 × 2²⁹ = $5 368 709.12. Offer A pays $30 000 in total; offer B pays more than $10 million in total. Exponential growth starts slowly and then overwhelms linear growth.

Your turn

1

Solve 7ˣ = 200 exactly and to three decimal places.

2

Solve 3^(x + 2) = 40 to three decimal places.

3

Solve 2(1.5)ˣ = 24 to three decimal places.

4

Solve 6ˣ = 5^(x + 1) to three decimal places.

5

Solve 4ˣ − 5(2ˣ) + 4 = 0.

6

Solve 2^(2x) − 2ˣ − 12 = 0.

7

How long does it take $500 at 4% per year compounded annually to grow to $1500?

More practice: warm-up

1

Take the log of both sides of 5ˣ = 9 and use the power law.

2

Solve 2ˣ = 10 to three decimal places.

3

Solve 10ˣ = 3 to three decimal places.

4

Solve 4ˣ = 17 to three decimal places.

5

What should you do first to solve 3(2ˣ) = 21?

More practice: on your own

1

Solve 5(3^(2x)) = 100 to three decimal places.

2

Solve 3^(2x) = 4^(x + 1) exactly, then to three decimal places.

3

Solve 25ˣ = 5^(x + 1) + 6. (Hint: 25ˣ = (5ˣ)².)

4

A town of 12 000 people grows by 2.5% a year. When will it reach 20 000?

5

Solve 2^(x − 1) = 5^(2x) to three decimal places.

Challenge

1

Solve 3ˣ × 2^(2x) = 100 to three decimal places.

2

Solve 2^(x + 2) + 2^(x + 1) + 2ˣ = 21 exactly.

3

Solve 6^(2x) − 7(6ˣ) + 6 = 0.

Check your understanding

Quiz

1.

After taking the common log of both sides of 6ˣ = 40, which equation do you get?

2.

To three decimal places, the solution of 2^(x + 1) = 15 is

3.

What is the exact solution of 5ˣ = 2^(x + 3)?

4.

Solving 9ˣ − 2(3ˣ) − 3 = 0 with u = 3ˣ gives u = 3 or u = −1. What are the solutions for x?

5.

$2000 is invested at 5% per year, compounded annually. After how many full years does it first exceed $3000?

6.

Which is a correct first step for 7(2ˣ) = 56?

For parents and teachers

Big idea (A3.2): every exponential equation can be solved by isolating the power and taking the logarithm of both sides; the power law brings the variable down. Look for a common base first, because it gives exact answers with no calculator.

How to help: Ask your teen to give both the exact answer (with logs) and the decimal, then check by substitution. For money and population problems, ask whether the answer makes sense in context (should it be rounded up to a whole year?).

Common difficulties:

  • Taking logs before isolating the power, then writing log (4 × 1.06ᵗ) = t log 4.24.
  • Losing brackets: (2x − 1) log 5, not 2x − 1 log 5.
  • Not collecting x-terms on one side when the variable appears in both exponents.
  • Keeping an impossible root such as 3ˣ = −1.