Solve equations of the form e^(kx) = a

About this lesson

Precalculus › Exponential and logarithmic functions

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Solve equations of the form e^(kx) = a

Warm-up

Three quick questions from earlier lessons.

Discover

A colony of bacteria doubles every hour, so after tt hours there are e0.69te^{0.69t} times as many. When will there be 7 times as many? Try to estimate before reading on: between which two whole hours?

Key idea

To undo exe^x, use its inverse function: ln⁡\ln.

ex=a  ⟺  x=ln⁡ae^x = a \iff x = \ln a, for a>0a > 0

Since ln⁡(ex)=x\ln(e^{x}) = x, taking ln⁡\ln of both sides "removes" the exponential.

Worked example

Solve e2x=7e^{2x} = 7.

2x=ln⁡72x = \ln 7
Apply ln to both sides: ln undoes e.
x=ln⁡72x = \dfrac{\ln 7}{2}
Divide both sides by 2.

Your turn, guided

Your turn: solve e3x=5e^{3x} = 5. Fill in the missing line.

3x=ln⁡53x = \ln 5
Apply ln to both sides.
x=ln⁡53x = \dfrac{\ln 5}{3}
Divide both sides by 3.

Practice ★

  1. Solve e5x=25e^{ 5x } = 25.
  2. Solve e5x=7e^{ 5x } = 7.
  3. Solve e5x=9e^{ 5x } = 9.
  4. Solve e3x=15e^{ 3x } = 15.
  5. Solve e5x=10e^{ 5x } = 10.

Watch out!

Practice ★★

  1. Solve e4x+4=12e^{ 4x + 4 } = 12.
  2. Solve e6x+5=37e^{ 6x + 5 } = 37.
  3. Solve e2x+7=16e^{ 2x + 7 } = 16.
  4. Solve e3x+6=27e^{ 3x + 6 } = 27.
  5. Solve e6x+7=39e^{ 6x + 7 } = 39.

Exit ticket

Three questions to check you have got it.

  1. Solve e4x+4=12e^{ 4x + 4 } = 12.
  2. Solve e6x+5=37e^{ 6x + 5 } = 37.
  3. Solve e2x+7=16e^{ 2x + 7 } = 16.

Summary card

Solving ekx=ae^{kx} = a (with a>0a > 0)

  1. Take ln⁡\ln of both sides: kx=ln⁡akx = \ln a.
  2. Divide by kk: x=ln⁡akx = \dfrac{\ln a}{k}.

Watch out: ee is not a factor. Never divide by ee.

One more step is optional: a challenge. Try it, or finish the lesson now.

Challenge ★★★

  1. Solve 3e4x+6=60 3e^{ 4x } + 6 = 60.
  2. Solve 5e2x+6=25 5e^{ 2x } + 6 = 25.
  3. Solve 4e2x+8=36 4e^{ 2x } + 8 = 36.

End of the lesson

Next lesson: Solve exponential equations in any base with logarithms Back to the chapter

Answers: Solve equations of the form e^(kx) = a

A

  1. ln⁡(25)5\frac{\ln{\left(25 \right)}}{5}
  2. ln⁡(7)5\frac{\ln{\left(7 \right)}}{5}
  3. ln⁡(9)5\frac{\ln{\left(9 \right)}}{5}
  4. ln⁡(15)3\frac{\ln{\left(15 \right)}}{3}
  5. ln⁡(10)5\frac{\ln{\left(10 \right)}}{5}

B

  1. −1+ln⁡(12)4-1 + \frac{\ln{\left(12 \right)}}{4}
  2. −56+ln⁡(37)6- \frac{5}{6} + \frac{\ln{\left(37 \right)}}{6}
  3. −72+ln⁡(16)2- \frac{7}{2} + \frac{\ln{\left(16 \right)}}{2}
  4. −2+ln⁡(27)3-2 + \frac{\ln{\left(27 \right)}}{3}
  5. −76+ln⁡(39)6- \frac{7}{6} + \frac{\ln{\left(39 \right)}}{6}

C

  1. −1+ln⁡(12)4-1 + \frac{\ln{\left(12 \right)}}{4}
  2. −56+ln⁡(37)6- \frac{5}{6} + \frac{\ln{\left(37 \right)}}{6}
  3. −72+ln⁡(16)2- \frac{7}{2} + \frac{\ln{\left(16 \right)}}{2}

D

  1. ln⁡(18)4\frac{\ln{\left(18 \right)}}{4}
  2. ln⁡(195)2\frac{\ln{\left(\frac{19}{5} \right)}}{2}
  3. ln⁡(7)2\frac{\ln{\left(7 \right)}}{2}

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