2.7: Semi-Log Plots | AP Precalculus | Code Scholars Tutoring

How it works

Taking the logarithm of positive outputs turns an exponential model into a linear relationship. A straight pattern on a semi-log plot supports exponential growth or decay. The slope depends on the logarithm base.

y=ab^x\ \Rightarrow\ \log_{10}y=\log_{10}a+x\log_{10}b

Worked examples

Example 1

For $\displaystyle y=10\cdot100^x$ , write the equation for $\displaystyle \log_{10}y$ .

It is $\displaystyle \log_{10}y=1+2x$ .

Example 2

A semi-log plot has equation $\displaystyle \log_{10}y=2+0.3x$ . Recover y.

Exponentiate: $\displaystyle y=10^{2+0.3x}=100(10^{0.3})^x$ .

Problems

Problem 2.7.1

At inputs $\displaystyle 0,1,2,3$ , outputs are $\displaystyle 1,10,100,1000$ . Find their base-10 logs and describe the semi-log pattern.

Points 0,0; 1,1; 2,2; 3,3 on axes x and log base 10 of y. — The original outputs are 1, 10, 100, and 1000.

Show worked solution

The log outputs are $\displaystyle 0,1,2,3$ , so the semi-log points lie on the line $\displaystyle \log_{10}y=x$ .

Points 0,0; 1,1; 2,2; 3,3 on axes x and log base 10 of y. The solution adds labeled relationships. — A straight line in these coordinates represents y = 10 to the power x.

Problem 2.7.2

Can a real semi-log plot use an output of $\displaystyle 0$ ?

Show worked solution

No. The logarithm of zero is undefined in the real numbers.

Problem 2.7.3

If $\displaystyle \ln y=3-0.2x$ , find the per-unit factor in the original model.

Show worked solution

Exponentiating gives $\displaystyle y=e^3(e^{-0.2})^x$ , with decay factor $\displaystyle e^{-0.2}\approx0.8187$ .

Lesson Problems PDF Solutions PDF Unit MCQ

How it works

Worked examples

Example 1

Example 2

Problems

Problem 2.7.1

Problem 2.7.2

Problem 2.7.3

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