Quadratic Exponential Equations

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Key Ideas Easy Medium Hard

Let u = b^x. Then b^2x = u².

u must be positive.

What to Remember

A quadratic in disguise

9^x is (3^x)². e^2x is (e^x)².

Let u equal the single exponential. Then you have a quadratic in u.

Keep only positive u

b^x and e^x are never 0 or negative.

If a factor gives u ≤ 0, that factor gives no solution.

Level 1

Easy

Both solutions are whole numbers

Example

Solve 9^x − 103^x + 9 = 0.

9^x − 103^x + 9 = 0

Let u = 3^x. Then 9^x = u².

u² − 10u + 9 = 0

Factor.

(u − 1)(u − 9) = 0

(u − 1) = 0

u = 1

3^x = 1

x = 0

(u − 9) = 0

u = 9

3^x = 9

x = 2

x = 0 or x = 2

Answer: x = 0 or x = 2

Solve 4^x − 62^x + 8 = 0.

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4^x − 62^x + 8 = 0

Let u = 2^x. Then 4^x = u².

u² − 6u + 8 = 0

Factor.

(u − 4)(u − 2) = 0

(u − 4) = 0

u = 4

2^x = 4

x = 2

(u − 2) = 0

u = 2

2^x = 2

x = 1

x = 2 or x = 1

Answer: x = 2 or x = 1

Solve 9^x − 43^x + 3 = 0.

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9^x − 43^x + 3 = 0

Let u = 3^x. Then 9^x = u².

u² − 4u + 3 = 0

Factor.

(u − 1)(u − 3) = 0

(u − 1) = 0

u = 1

3^x = 1

x = 0

(u − 3) = 0

u = 3

3^x = 3

x = 1

x = 0 or x = 1

Answer: x = 0 or x = 1

Solve 25^x − 65^x + 5 = 0.

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25^x − 65^x + 5 = 0

Let u = 5^x. Then 25^x = u².

u² − 6u + 5 = 0

Factor.

(u − 1)(u − 5) = 0

(u − 1) = 0

u = 1

5^x = 1

x = 0

(u − 5) = 0

u = 5

5^x = 5

x = 1

x = 0 or x = 1

Answer: x = 0 or x = 1

Solve 4^x − 32^x + 2 = 0.

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4^x − 32^x + 2 = 0

Let u = 2^x. Then 4^x = u².

u² − 3u + 2 = 0

Factor.

(u − 1)(u − 2) = 0

(u − 1) = 0

u = 1

2^x = 1

x = 0

(u − 2) = 0

u = 2

2^x = 2

x = 1

x = 0 or x = 1

Answer: x = 0 or x = 1

Solve 9^x − 123^x + 27 = 0.

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9^x − 123^x + 27 = 0

Let u = 3^x. Then 9^x = u².

u² − 12u + 27 = 0

Factor.

(u − 3)(u − 9) = 0

(u − 3) = 0

u = 3

3^x = 3

x = 1

(u − 9) = 0

u = 9

3^x = 9

x = 2

x = 1 or x = 2

Answer: x = 1 or x = 2

Level 2

Medium

A root that has to be rejected

Example

Solve e^2x − 3e^x − 4 = 0.

e^2x − 3e^x − 4 = 0

Let u = e^x. Then e^2x = u².

u² − 3u − 4 = 0

Factor.

(u − 4)(u + 1) = 0

(u − 4) = 0

u = 4

e^x = 4

x = ln 4

(u + 1) = 0

u = −1

e^x = −1

No solution, because e^x is always positive.

x = ln 4

Answer: x = ln 4

Solve e^2x − 5e^x + 6 = 0.

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e^2x − 5e^x + 6 = 0

Let u = e^x. Then e^2x = u².

u² − 5u + 6 = 0

Factor.

(u − 2)(u − 3) = 0

(u − 2) = 0

u = 2

e^x = 2

x = ln 2

(u − 3) = 0

u = 3

e^x = 3

x = ln 3

x = ln 2 or x = ln 3

Answer: x = ln 2 or x = ln 3

Solve 2e^2x + e^x − 6 = 0.

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2e^2x + e^x − 6 = 0

Let u = e^x. Then e^2x = u².

2u² + u − 6 = 0

Factor.

(2u − 3)(u + 2) = 0

(2u − 3) = 0

u = 32

e^x = 32

x = ln(32)

(u + 2) = 0

u = −2

e^x = −2

No solution, because e^x is always positive.

x = ln(32)

Answer: x = ln(32)

Solve 34^x − 72^x − 6 = 0.

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34^x − 72^x − 6 = 0

Let u = 2^x. Then 4^x = u².

3u² − 7u − 6 = 0

Factor.

(3u + 2)(u − 3) = 0

(3u + 2) = 0

u = −23

2^x = −23

No solution, because 2^x is always positive.

(u − 3) = 0

u = 3

2^x = 3

x = log₂ 3

Answer: x = log₂ 3

Solve e^2x + 4e^x − 5 = 0.

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e^2x + 4e^x − 5 = 0

Let u = e^x. Then e^2x = u².

u² + 4u − 5 = 0

Factor.

(u + 5)(u − 1) = 0

(u + 5) = 0

u = −5

e^x = −5

No solution, because e^x is always positive.

(u − 1) = 0

u = 1

e^x = 1

x = 0

Answer: x = 0

Solve 29^x − 33^x − 2 = 0.

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29^x − 33^x − 2 = 0

Let u = 3^x. Then 9^x = u².

2u² − 3u − 2 = 0

Factor.

(2u + 1)(u − 2) = 0

(2u + 1) = 0

u = −12

3^x = −12

No solution, because 3^x is always positive.

(u − 2) = 0

u = 2

3^x = 2

x = log₃ 2

Answer: x = log₃ 2

Level 3

Hard

A fraction that stays in the answer, or two natural logs

Example

Solve 9^x − 43^x − 5 = 0.

9^x − 43^x − 5 = 0

Let u = 3^x. Then 9^x = u².

u² − 4u − 5 = 0

Factor.

(u − 5)(u + 1) = 0

(u − 5) = 0

u = 5

3^x = 5

x = log₃ 5

(u + 1) = 0

u = −1

3^x = −1

No solution, because 3^x is always positive.

x = log₃ 5

Answer: x = log₃ 5

Solve 49^x − 83^x − 5 = 0.

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49^x − 83^x − 5 = 0

Let u = 3^x. Then 9^x = u².

4u² − 8u − 5 = 0

Factor.

(2u − 5)(2u + 1) = 0

(2u − 5) = 0

u = 52

3^x = 52

x = log₃ (52)

(2u + 1) = 0

u = −12

3^x = −12

No solution, because 3^x is always positive.

x = log₃ (52)

Answer: x = log₃ (52)

Solve e^2x − 6e^x + 8 = 0.

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e^2x − 6e^x + 8 = 0

Let u = e^x. Then e^2x = u².

u² − 6u + 8 = 0

Factor.

(u − 2)(u − 4) = 0

(u − 2) = 0

u = 2

e^x = 2

x = ln 2

(u − 4) = 0

u = 4

e^x = 4

x = ln 4

x = ln 2 or x = ln 4

Answer: x = ln 2 or x = ln 4

Solve 25^x − 25^x − 8 = 0.

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25^x − 25^x − 8 = 0

Let u = 5^x. Then 25^x = u².

u² − 2u − 8 = 0

Factor.

(u − 4)(u + 2) = 0

(u − 4) = 0

u = 4

5^x = 4

x = log₅ 4

(u + 2) = 0

u = −2

5^x = −2

No solution, because 5^x is always positive.

x = log₅ 4

Answer: x = log₅ 4

Solve 34^x − 52^x − 2 = 0.

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34^x − 52^x − 2 = 0

Let u = 2^x. Then 4^x = u².

3u² − 5u − 2 = 0

Factor.

(3u + 1)(u − 2) = 0

(3u + 1) = 0

u = −13

2^x = −13

No solution, because 2^x is always positive.

(u − 2) = 0

u = 2

2^x = 2

x = 1

Answer: x = 1

Solve 2e^2x − 9e^x + 4 = 0.

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2e^2x − 9e^x + 4 = 0

Let u = e^x. Then e^2x = u².

2u² − 9u + 4 = 0

Factor.

(2u − 1)(u − 4) = 0

(2u − 1) = 0

u = 12

e^x = 12

x = ln(12)

x = −ln 2

(u − 4) = 0

u = 4

e^x = 4

x = ln 4

x = −ln 2 or x = ln 4

Answer: x = −ln 2 or x = ln 4

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