Precalculus Exponential and Logarithmic Equations Lesson (Free PDF)

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This exp log equations lesson for Precalculus covers exponential and logarithmic equations (Common Core F-LE.A.4, F-BF.B.5, A-REI.D.11), with a free PDF to print.

To solve an exponential equation, isolate the power, then either match the bases or take a logarithm of both sides and use the power rule to bring the exponent down. To solve a logarithmic equation, combine the logs, rewrite in exponential form and solve. Always check logarithmic solutions, because an argument that is zero or negative makes a solution extraneous.

Precalculus · Common Core State Standards: F-LE.A.4, F-BF.B.5, A-REI.D.11

Key facts in this exp log equations lesson

Key facts
  • Same base: if b^m = b^n then m = n.
  • Different bases: take ln of both sides, then x ln b = ln c.
  • log_b u = c is equivalent to u = b^c.
  • Check log solutions: arguments must be positive; reject extraneous values.
  • Continuous doubling time is ln 2 divided by the rate r.

How do you solve an exponential equation?

If both sides can be written with the same base, set the exponents equal: \(9^{x}=27\) becomes \(3^{2x}=3^{3}\), so \(2x=3\) and \(x=\frac32\). When the bases cannot be matched, take a logarithm of both sides. The power rule brings the exponent down: \(3^{x}=20\) gives \(x\ln3=\ln20\), so \(x=\frac{\ln20}{\ln3}\approx2.727\). Always isolate the exponential part first.

1. Isolatethe powera b^x = c2. Takeln ofboth sides3. Power rulex ln b =ln(c/a)4. Dividex = ln(c/a)/ ln b
Four steps for an equation of the form a b^x = c.

For equations with base \(e\), use the natural log, since \(\ln e^{u}=u\). For example, \(e^{2x}=15\) gives \(2x=\ln15\), so \(x=\frac{\ln15}{2}\approx1.354\).

How do you solve a logarithmic equation?

  1. Combine all logarithms into one using the product, quotient and power rules.
  2. Rewrite in exponential form: \(\log_{b}u=c\) means \(u=b^{c}\).
  3. Solve the resulting equation.
  4. Check each answer in the original equation, because arguments must be positive.

Example: Solve \(\log x+\log(x-3)=1\). Combine: \(\log\left(x(x-3)\right)=1\), so \(x^{2}-3x=10\). Then \(x^{2}-3x-10=(x-5)(x+2)=0\). The value \(x=-2\) makes \(\log(-2)\) undefined, so it is extraneous. The solution is \(x=5\).

123456789-2-112xy(5, 1)
The curve \(y=\log x+\log(x-3)\) exists only for \(x>3\) and meets \(y=1\) once, at \(x=5\).

Why can you solve an equation by graphing?

Every solution of \(f(x)=g(x)\) is the \(x\)-coordinate of an intersection point of \(y=f(x)\) and \(y=g(x)\). Graphing is a good check and works even when algebra is hard. The graph of \(y=2^{x}\) crosses \(y=10\) at \(x=\log_{2}10\approx3.32\).

-112345481216xy(3.32, 10)
The solution of \(2^{x}=10\) is the \(x\)-value where the curve meets the line.

How are these equations used in applications?

Doubling time, half-life and time to reach a savings goal all require solving for an exponent. With continuous growth \(A=Pe^{rt}\), the doubling time satisfies \(e^{rt}=2\), so \(t=\frac{\ln2}{r}\). At \(6\%\) that is \(\frac{0.6931}{0.06}\approx11.55\) years, regardless of the starting amount.

5101520100020003000t (yr)A(11.55, 2000)
Money growing continuously at 6 percent doubles in about 11.55 years.

When you report an answer from a real situation, round sensibly and include units, and check that it is reasonable: a result of 11.55 years for doubling at 6% matches the rule of thumb that 72 divided by the percent rate estimates the doubling time.

Exp log equations lesson: questions and answers

When can I just set exponents equal?

When both sides can be written as a power of the same base, such as 2^x = 32 = 2^5. Then x = 5.

Which logarithm should I use to solve 3^x = 20?

Any base works, as long as you use the same one on both sides. Natural log or common log are the easiest to compute: x = ln 20 / ln 3 = log 20 / log 3.

What is an extraneous solution in a log equation?

It is an answer that satisfies the algebra but makes the argument of a logarithm zero or negative in the original equation, so it must be rejected.

How do I find a doubling time?

Set the growth model equal to twice the starting amount and solve for t. For continuous growth at rate r, t = ln 2 / r.

Can I solve these equations by graphing?

Yes. Graph each side as a function and find the x-coordinate of the intersection. It is a good way to check an algebraic answer.

How to use this exp log equations lesson

Read the definitions first, then follow the worked method line by line and redo the example on paper without looking. When every step makes sense, move on to the worksheet, then check yourself with the quiz.

Download the exp log equations lesson as a PDF with the button below and keep it with your notes.

Download the exp log equations lesson as a PDF with the button below and keep it with your notes.

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