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Precalculus Honors · Study guide

1.4 · Exponential and logarithmic functions and models

Use inverse relationships to solve equations, enforce logarithmic domains, and interpret growth rates and doubling times.

Precalculus Honors · Reading: Demana 10th ed. §§3.1–3.6; Stewart 8th ed. §§4.1–4.6

Learning objectives

  • Solve exponential and logarithmic equations with domain checks.
  • Interpret exponential growth and decay models.
  • Connect exponential and logarithmic graphs as inverses.

An exponential has a constant multiplicative ratio

For P(t)=P0bt\displaystyle P(t)=P_0b^t, a unit increase in time multiplies the amount by b\displaystyle b. If b>1\displaystyle b>1 there is growth; if 0<b<1\displaystyle 0<b<1 there is decay. The base cannot be zero or negative for a real exponential defined on all real inputs.

The continuous model P0ekt\displaystyle P_0e^{kt} uses a continuous relative rate k\displaystyle k. A discrete increase of r\displaystyle r per period uses b=1+r\displaystyle b=1+r; the corresponding continuous coefficient is ln⁡(1+r)\displaystyle \ln(1+r), not necessarily r\displaystyle r.

A logarithm asks for an exponent

The statement log⁡bx=y\displaystyle \log_bx=y is equivalent to by=x\displaystyle b^y=x, for b>0\displaystyle b>0, b≠1\displaystyle b\ne1, and x>0\displaystyle x>0. Exponential and logarithmic graphs reflect across y=x\displaystyle y=x when their bases match.

The product, quotient, and power laws follow from exponent rules. They do not turn a logarithm of a sum into a sum of logarithms. In an equation, check every original log argument for positivity.

log⁡b(uv)=log⁡bu+log⁡bv\log_b(uv)=\log_bu+\log_bv

Solve models with logarithms

For doubling time T\displaystyle T, the continuous coefficient satisfies ekT=2\displaystyle e^{kT}=2, so k=ln⁡2/T\displaystyle k=\ln2/T. For half-life T\displaystyle T, use k=−ln⁡2/T\displaystyle k=-\ln2/T. Retain exact logarithms until a decimal is required.

A logarithmic domain may remove an algebraic candidate. For log⁡2x+log⁡2(x−2)=3\displaystyle \log_2x+\log_2(x-2)=3, require x>2\displaystyle x>2 before solving x(x−2)=8\displaystyle x(x-2)=8.

Worked examples

A logarithmic equation

Solve log⁡2x+log⁡2(x−2)=3\displaystyle \log_2x+\log_2(x-2)=3.

  1. The domain is x>2\displaystyle x>2.
  2. Combine the logs to obtain x(x−2)=8\displaystyle x(x-2)=8.
  3. The candidates are 4\displaystyle 4 and −2\displaystyle -2; only 4\displaystyle 4 is allowed.

Doubling time

An amount starts at 50\displaystyle 50 and doubles every 3\displaystyle 3 years. Find the amount after 6\displaystyle 6 years.

  1. Use P(t)=50⋅2t/3\displaystyle P(t)=50\cdot2^{t/3}.
  2. At 6\displaystyle 6 years, the exponent is 2\displaystyle 2.
  3. The amount is 200\displaystyle 200.

Common mistakes

  • Using logarithm rules on a sum.
  • Keeping solutions with nonpositive original log arguments.
  • Confusing additive growth with constant percentage growth.

Try it yourself

No calculator required.

1. Compare f(x)=2x\displaystyle f(x)=2^x and g(x)=log⁡2x\displaystyle g(x)=\log_2x. State domains and their relationship.

The graphs 2 to the x and log base 2 of x reflect across y = x.
Compare the input and output roles on the two curves.
Show worked solution

The exponential domain is all real numbers and its range is (0,∞)\displaystyle (0,\infty). The logarithm reverses these. They are inverse functions and their graphs reflect across y=x\displaystyle y=x.

The graphs 2 to the x and log base 2 of x reflect across y = x. The domains and ranges swap because the functions are inverses.
The domains and ranges swap because the functions are inverses.

2. Solve log⁡2x+log⁡2(x−2)=3\displaystyle \log_2x+\log_2(x-2)=3.

Show worked solution

The original domain is x>2\displaystyle x>2. Combining logs gives x2−2x−8=0\displaystyle x^2-2x-8=0, with candidates 4\displaystyle 4 and −2\displaystyle -2. Only x=4\displaystyle x=4 is valid.

3. An amount starts at 80\displaystyle 80 and halves every 4\displaystyle 4 years. Write its model and evaluate it at 12\displaystyle 12 years.

Show worked solution

The model is P(t)=80(1/2)t/4\displaystyle P(t)=80(1/2)^{t/4}. At 12\displaystyle 12 years, three half-lives give P(12)=10\displaystyle P(12)=10.

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