Exponential and Logarithmic Functions and Models: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Exponential Log Worksheet Answer Key: problems 1–5

Problem 1: Exponential to logarithmic form

1. \(\log_{2} 128=7\)
2. \(\log 0.001=-3\)
3. \(\log_{9} 3=\frac{1}{2}\)
4. \(\ln 1=0\)

Problem 2: Logarithmic to exponential form

1. \(5^{3}=125\)
2. \(4^{-3}=\frac{1}{64}\)
3. \(36^{1/2}=6\)
4. \(e^{t}=20\)

Problem 3: Evaluate logarithms

1. \(3^{4}=81\), so 4
2. \(2^{-5}=\frac{1}{32}\), so -5
3. \(25^{1/2}=5\), so \(\frac{1}{2}\)
4. \((\frac{1}{4})^{-2}=16\), so -2
5. \(\sqrt{e}=e^{1/2}\), so \(\frac{1}{2}\)

Problem 4: Points on an inverse graph

1. Swap the coordinates: \(A^{\prime}(\frac{1}{2},-1),\ B^{\prime}(1,0),\ C^{\prime}(4,2),\ D^{\prime}(8,3)\)
2. \(f^{-1}(x)=\log_{2} x\)
3. The domain of the inverse is the range of \(f\), so \(x > 0\)

Problem 5: Finding inverse functions

1. \(y=4^{x}\) becomes \(x=4^{y}\), so \(f^{-1}(x)=\log_{4} x\)
2. \(x=2\cdot 3^{y}\) gives \(3^{y}=\frac{x}{2}\), so \(f^{-1}(x)=\log_{3}\frac{x}{2}\)
3. \(x=\ln (y+3)\) gives \(y+3=e^{x}\), so \(f^{-1}(x)=e^{x}-3\)
4. \(x-1=\log_{2} y\) gives \(y=2^{x-1}\), so \(f^{-1}(x)=2^{x-1}\)

Precalculus Exponential Log Worksheet Answer Key: problems 6–10

Problem 6: Verify inverses by composition

\(f(g(x))=e^{2\cdot\frac{1}{2}\ln x}=e^{\ln x}=x\) for \(x > 0\).

\(g(f(x))=\frac{1}{2}\ln e^{2x}=\frac{1}{2}(2x)=x\) for all \(x\).

Both compositions return \(x\), so the functions are inverses.

Problem 7: Features of a log graph

1. The asymptote is \(x=2\).
2. We need \(x-2 > 0\), so the domain is \(x > 2\) and the range is all real numbers.
3. \(\log_{3}(x-2)=-1\) gives \(x-2=3^{-1}=\frac{1}{3}\), so \(x=\frac{7}{3}\)
4. \(g(29)=\log_{3} 27+1=3+1=\)4

Problem 8: Expand logarithms

1. \(3\log_{b} x+2\log_{b} y\)
2. \(\ln 5+\ln x-4\ln y\)
3. \(\frac{1}{2}(\log_{2}8+\log_{2}x-\log_{2}z)=\frac{3}{2}+\frac{1}{2}\log_{2}x-\frac{1}{2}\log_{2}z\)

Problem 9: Condense logarithms

1. \(\log 3^{2}+\log 5=\)\(\log 45\)
2. \(\ln\frac{12}{4}+\ln x^{2}=\)\(\ln (3x^{2})\)
3. \(\log_{2}\sqrt{x}+\log_{2}y^{2}-\log_{2}z^{3}=\)\(\log_{2}\frac{\sqrt{x}\,y^{2}}{z^{3}}\)

Problem 10: Using given logarithm values

1. \(\log (2\cdot 3)=0.3010+0.4771=\)0.7781
2. \(\log (2\cdot 3^{2})=0.3010+2(0.4771)=\)1.2552
3. \(\log\frac{3}{2}=0.4771-0.3010=\)0.1761
4. \(\log 2^{3}=3(0.3010)=\)0.9030
5. \(\log 2-2\log 3=0.3010-0.9542=\)-0.6532

Precalculus Exponential Log Worksheet Answer Key: problems 11–15

Problem 11: Change of base

1. \(\frac{\ln 40}{\ln 5}\approx\frac{3.6889}{1.6094}=\)2.292
2. \(\frac{\ln 3}{\ln 7}\approx\frac{1.0986}{1.9459}=\)0.565
3. \(\frac{\ln 100}{\ln 2}\approx\frac{4.6052}{0.6931}=\)6.644

Problem 12: Same-base exponential equations

1. \(2^{2x+2}=2^{5}\), so \(2x+2=5\) and \(x=\frac{3}{2}\)
2. \(3^{2x}=3^{3x-6}\), so \(2x=3x-6\) and \(x=6\)
3. \(2^{-x}=2^{4}\), so \(x=-4\)
4. \(5^{2x-2}=5^{3x}\), so \(2x-2=3x\) and \(x=-2\)

Problem 13: Solve using logarithms

1. \(2^{x}=20\), so \(x=\frac{\ln 20}{\ln 2}\approx\) 4.322
2. \(x-1=\frac{\ln 50}{\ln 7}\), so \(x=1+\frac{\ln 50}{\ln 7}\approx\) 3.010
3. \(e^{2x}=9\), so \(x=\frac{\ln 9}{2}=\ln 3\approx\) 1.099
4. \(0.9^{x}=0.25\), so \(x=\frac{\ln 0.25}{\ln 0.9}\approx\) 13.158

Problem 14: Estimate, then solve

1. Draw the line \(y=40\): it meets the curve near \(x\approx 5.2\).
2. \(e^{0.4x}=8\), so \(0.4x=\ln 8\) and \(x=\frac{\ln 8}{0.4}\approx\) 5.199.

Problem 15: Logarithmic equations

1. \(\log_{2}(x^{2}-9)=4\), so \(x^{2}-9=16\) and \(x=\pm 5\). The value \(x=-5\) makes \(x+3\) negative, so the solution is \(x=5\)
2. \(x(x-2)=8\) gives \(x^{2}-2x-8=0\), so \((x-4)(x+2)=0\). Reject \(x=-2\): \(x=4\)
3. \(2x-1=3^{2}=9\), so \(x=5\)
4. \(x+6=x^{2}\) gives \((x-3)(x+2)=0\). Reject \(x=-2\): \(x=3\)

Precalculus Exponential Log Worksheet Answer Key: problems 16–20

Problem 16: Quadratic form

1. Let \(u=e^{x}\): \(u^{2}-5u+6=0\), so \(u=2\) or \(u=3\). Then \(x=\ln 2\approx 0.693\) or \(x=\ln 3\approx 1.099\)
2. Let \(u=2^{x}\): \(u^{2}-6u+8=0\), so \(u=2\) or \(u=4\). Then \(x=1\) or \(x=2\)

Problem 17: Transformations of an exponential

1. Shift right 1 unit, reflect over the x-axis, then shift up 3 units.
2. The asymptote moves up 3: \(y=3\)
3. \(h(0)=-2^{-1}+3=\)\(\frac{5}{2}\)
4. \(2^{x-1}=3\), so \(x=1+\log_{2}3\approx 2.585\): \(x=1+\log_{2}3\)
5. The graph lies below its asymptote: \(y < 3\)

Problem 18: Rewrite growth and decay models

1. \(0.8^{t}=e^{t\ln 0.8}\), so \(A=200e^{kt}\) with \(k=\ln 0.8\approx -0.2231\). Negative \(k\) means decay.
2. \(e^{0.07}\approx 1.0725\), so the growth is about 7.25% per unit, a bit more than 7%.
3. \(2^{t/3}=e^{(t/3)\ln 2}\), so \(k=\frac{\ln 2}{3}\approx 0.2310\)

Problem 19: Compound interest time

\(3000=2000(1.01)^{4t}\), so \(1.01^{4t}=1.5\).

\(4t=\frac{\ln 1.5}{\ln 1.01}\approx\frac{0.40547}{0.00995}=40.75\)

\(t\approx\) 10.19 years.

Problem 20: Continuous compounding

\(10000=5000e^{0.035t}\) gives \(e^{0.035t}=2\), so \(t=\frac{\ln 2}{0.035}\approx\) 19.80 years.

\(e^{0.035t}=2.4\), so \(t=\frac{\ln 2.4}{0.035}\approx\) 25.01 years.

Precalculus Exponential Log Worksheet Answer Key: problems 21–24

Problem 21: Radioactive decay dating

\(0.3=(\frac{1}{2})^{t/5730}\).

Take natural logs: \(\frac{t}{5730}\ln 0.5=\ln 0.3\), so \(t=5730\cdot\frac{\ln 0.3}{\ln 0.5}\approx 5730(1.7370)\)

The bone is about 9953 years old.

Problem 22: Logistic population model

1. \(P(0)=\frac{1200}{1+19}=\)60 students
2. 1,200 students, the numerator \(K\)
3. \(1+19e^{-0.5t}=2\), so \(e^{-0.5t}=\frac{1}{19}\) and \(t=2\ln 19\approx\) 5.89 hours
4. \(P(8)=\frac{1200}{1+19e^{-4}}\approx\frac{1200}{1.348}\approx\) 890 students

Problem 23: pH calculations

1. \(-\log (3.2\times 10^{-5})=-(0.5051-5)\approx\) 4.49
2. \([H^{+}]=10^{-8.2}\approx\) \(6.31\times 10^{-9}\) mol/L
3. \(10^{-3}\) versus \(10^{-6}\): the ratio is \(10^{3}=\)1,000 times

Problem 24: Decibels and earthquakes

1. \(90-60=30\) dB is \(3\) steps of 10 dB, so the intensity ratio is \(10^{3}=\)1,000
2. \(110=10\log\frac{I}{I_{0}}\) gives \(\log\frac{I}{I_{0}}=11\), so \(\frac{I}{I_{0}}=\)\(1\times 10^{11}\)
3. The difference is \(7.1-5.4=1.7\), so the factor is \(10^{1.7}\approx 50.12\), about 50.

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