Solving Exponential Equations

Using logarithms to find unknown exponents

SPEC 6.5 • EXPONENTIALS AND LOGARITHMS

Why Use Logarithms?

Consider the equation

2^{x} = 10

. How do we find

x

?

The Problem

We need to "undo" the exponential. Since logarithms are the inverse of exponentials, we use logs to isolate the unknown exponent.

Key Principle

If

a^{x} = b

, then taking logarithms of both sides:

lo g (a^{x}) = lo g (b)

x lo g a = lo g b

x = \frac{l o g b}{l o g a}

Note

You can use any base logarithm (log, ln, log₂, etc.) — the result will be the same. We typically use log (base 10) because it's the main button on most calculators.

The Basic Method

To solve

a^{x} = b

where

a > 0

,

a \neq = 1

, and

b > 0

:

Method

Step 1: Take log of both sides

Step 2: Use the power law:

lo g (a^{x}) = x lo g a

Step 3: Solve for

x

a^{x} = b ⟹ x = \frac{l o g b}{l o g a}

Important

The equation

a^{x} = b

only has a solution when

b > 0

.

For example,

2^{x} = - 4

has NO SOLUTION because

2^{x} > 0

for all real

x

.

Worked Example01

Solve $5^{x} = 12$ , giving your answer to 3 significant figures.

•

1. Take log of both sides:

lo g (5^{x}) = lo g (12)

•

2. Use power law:

x lo g 5 = lo g 12

•

3. Solve for

x

:

x = \frac{l o g 12}{l o g 5}

•

4. Calculate:

x = \frac{1.0792...}{0.6990...} = 1.5439...

Answer: $x = 1.54$ (3 s.f.)

Exact answer: $x = \frac{l o g 12}{l o g 5}$

Equations with eˣ

Equations involving

e^{x}

are particularly straightforward because

ln

and

e^{x}

are inverses.

Direct Method

e^{x} = k ⟹ x = ln k

Simple: eˣ = k

e^{x} = 5

x = ln 5

With Coefficient: e^(ax) = k

e^{2 x} = 7

2 x = ln 7

x = \frac{l n 7}{2}

Remember

For

e^{x}

equations, use

ln

since it's the natural inverse. Taking ln "undoes" the exponential immediately:

ln (e^{something}) = something

Worked Example02

Solve $e^{2 x} = 7$ , giving your answer in the form $\frac{l n a}{b}$ .

•

1. Take ln of both sides:

ln (e^{2 x}) = ln 7

•

2. Since

ln (e^{2 x}) = 2 x

:

2 x = ln 7

•

3. Divide by 2:

x = \frac{l n 7}{2}

Answer: $x = \frac{l n 7}{2}$

Linear Expressions in Exponent

When the exponent is a linear expression like

a x + b

, we solve for

x

after taking log.

Method

e^{a x + b} = k

Step 1: Take ln:

a x + b = ln k

Step 2: Rearrange:

a x = ln k - b

Step 3: Solve:

x = \frac{l n k - b}{a}

Same Pattern for Other Bases

For

a^{p x + q} = k

:

(p x + q) lo g a = lo g k

x = \frac{l o g k - q l o g a}{p l o g a}

Worked Example03

Solve $e^{3 x - 1} = 20$ , giving your answer to 3 decimal places.

•

1. Take ln of both sides:

ln (e^{3 x - 1}) = ln 20

•

2. Simplify left side:

3 x - 1 = ln 20

•

3. Add 1:

3 x = ln 20 + 1

•

4. Divide by 3:

x = \frac{l n 20 + 1}{3}

•

5. Calculate:

x = \frac{2.9957... + 1}{3} = \frac{3.9957...}{3} = 1.3319...

Answer: $x = 1.332$ (3 d.p.)

Different Bases on Each Side

When you have different bases like

2^{x + 3} = 3^{2 x - 1}

, take log of both sides.

Method

Step 1: Take log of both sides

Step 2: Apply power law to each side

Step 3: Expand and collect

x

terms

Step 4: Factorise and solve for

x

Key Technique

After applying the power law, you'll have a linear equation in

x

. Collect all

x

terms on one side and factorise.

Worked Example04

Solve $2^{x + 3} = 3^{2 x - 1}$ , giving your answer to 3 significant figures.

•

1. Take log of both sides:

lo g (2^{x + 3}) = lo g (3^{2 x - 1})

•

2. Apply power law:

(x + 3) lo g 2 = (2 x - 1) lo g 3

•

3. Expand:

x lo g 2 + 3 lo g 2 = 2 x lo g 3 - lo g 3

•

4. Collect

x

terms:

x lo g 2 - 2 x lo g 3 = - lo g 3 - 3 lo g 2

•

5. Factorise:

x (lo g 2 - 2 lo g 3) = - lo g 3 - 3 lo g 2

•

6. Solve:

x = \frac{- l o g 3 - 3 l o g 2}{l o g 2 - 2 l o g 3}

•

7. Calculate:

x = \frac{- 0.4771... - 0.9031...}{0.3010... - 0.9542...} = \frac{- 1.380...}{- 0.6532...} = 2.113...

Answer: $x = 2.11$ (3 s.f.)

Disguised Quadratics

Some exponential equations are quadratics in disguise. Look for patterns like:

Pattern 1

e^{2 x} - 5 e^{x} + 6 = 0

Let

y = e^{x}

, then

e^{2 x} = (e^{x})^{2} = y^{2}

Equation becomes:

y^{2} - 5 y + 6 = 0

Pattern 2

4^{x} - 6 (2^{x}) + 8 = 0

Note:

4^{x} = (2^{2})^{x} = (2^{x})^{2}

Let

y = 2^{x}

, becomes:

y^{2} - 6 y + 8 = 0

Critical Check

After solving the quadratic, remember:

e^{x} > 0

always, so reject

y \leq 0

2^{x} > 0

always, so reject

y \leq 0

Worked Example05

Solve $e^{2 x} - 6 e^{x} + 8 = 0$ , giving exact answers.

•

1. Let

y = e^{x}

, so

e^{2 x} = y^{2}

•

2. Equation becomes:

y^{2} - 6 y + 8 = 0

•

3. Factorise:

(y - 2) (y - 4) = 0

•

4. So

y = 2

or

y = 4

•

5. Since

y = e^{x}

:

e^{x} = 2

or

e^{x} = 4

•

6. Both values are positive, so both are valid.

•

7. Take ln:

x = ln 2

or

x = ln 4

Answer: $x = ln 2$ or $x = ln 4$ (which equals $2 ln 2$ )

Worked Example06

Solve $4^{x} - 6 (2^{x}) + 8 = 0$ .

•

1. Recognise that

4^{x} = (2^{2})^{x} = (2^{x})^{2}

•

2. Let

y = 2^{x}

, so

4^{x} = y^{2}

•

3. Equation becomes:

y^{2} - 6 y + 8 = 0

•

4. Factorise:

(y - 2) (y - 4) = 0

•

5. So

y = 2

or

y = 4

•

6. Since

y = 2^{x}

:

2^{x} = 2

or

2^{x} = 4

•

7. Therefore:

x = 1

or

x = 2

Answer: $x = 1$ or $x = 2$

Using Index Laws to Form Quadratics

Some equations need index law manipulation before you can see the quadratic structure.

Example Pattern

3^{2 x} - 4 \cdot 3^{x + 1} + 27 = 0

Rewrite

3^{x + 1}

using index laws:

3^{x + 1} = 3^{x} \cdot 3 = 3 \cdot 3^{x}

So equation becomes:

3^{2 x} - 4 (3 \cdot 3^{x}) + 27 = 0

3^{2 x} - 12 \cdot 3^{x} + 27 = 0

Now let

y = 3^{x}

to get:

y^{2} - 12 y + 27 = 0

Key Technique

Use

a^{x + n} = a^{n} \cdot a^{x}

to simplify coefficients, then substitute to form a quadratic.

Worked Example07

Solve $3^{2 x} - 4 \cdot 3^{x + 1} + 27 = 0$ .

•

1. Rewrite

3^{x + 1} = 3 \cdot 3^{x}

:

3^{2 x} - 4 (3 \cdot 3^{x}) + 27 = 0

•

2. Simplify:

3^{2 x} - 12 \cdot 3^{x} + 27 = 0

•

3. Let

y = 3^{x}

, so

3^{2 x} = y^{2}

:

y^{2} - 12 y + 27 = 0

•

4. Factorise:

(y - 3) (y - 9) = 0

•

5. So

y = 3

or

y = 9

(both positive ✓)

•

6.

3^{x} = 3 \Rightarrow x = 1

•

7.

3^{x} = 9 = 3^{2} \Rightarrow x = 2

Answer: $x = 1$ or $x = 2$

Practice

Question 1: Solve

3^{x} = 20

, giving your answer to 3 significant figures.

•

1. Take log of both sides:

lo g (3^{x}) = lo g 20

•

2. Use power law:

x lo g 3 = lo g 20

•

3. Solve:

x = \frac{l o g 20}{l o g 3} = \frac{1.3010...}{0.4771...} = 2.727...

Answer: $x = 2.73$ (3 s.f.)

Practice

Question 2: Solve

e^{4 x - 2} = 15

, giving your answer to 3 decimal places.

•

1. Take ln:

4 x - 2 = ln 15

•

2. Add 2:

4 x = ln 15 + 2

•

3. Divide by 4:

x = \frac{l n 15 + 2}{4}

•

4. Calculate:

x = \frac{2.7080... + 2}{4} = \frac{4.7080...}{4} = 1.177...

Answer: $x = 1.177$ (3 d.p.)

Practice

Question 3: Solve

e^{2 x} - 7 e^{x} + 12 = 0

, giving exact answers.

•

1. Let

y = e^{x}

, so

e^{2 x} = y^{2}

•

2. Equation becomes:

y^{2} - 7 y + 12 = 0

•

3. Factorise:

(y - 3) (y - 4) = 0

•

4. So

y = 3

or

y = 4

(both positive ✓)

•

5.

e^{x} = 3

→

x = ln 3

•

6.

e^{x} = 4

→

x = ln 4 = 2 ln 2

Answer: $x = ln 3$ or $x = ln 4$

Summary

Key Methods

Equation TypeMethod

a^{x} = b

x = \frac{l o g b}{l o g a}

e^{a x + b} = k

Take ln:

a x + b = ln k

, then solve

a^{f (x)} = b^{g (x)}

Take log of both, apply power law, collect

x

terms

Quadratic in

e^{x}

Let

y = e^{x}

, solve quadratic, check

y > 0

4^{x}, 2^{x}

mixedUse

4^{x} = (2^{x})^{2}

, let

y = 2^{x}

Remember

Exponentials are always positive:

a^{x} > 0

for all real

x

Always check validity of solutions from quadratics

Give exact answers unless asked for decimals