Algebra 2 Logarithms Answer Key – Worked Solutions for 23 Problems

Algebra 2 Logarithms Worksheet Answer Key: problems 1–5

1. Exponential to log form

The base stays the base and the exponent becomes the value of the log.

1. \(\log_{2} 32=5\)
2. \(\log_{10} 1000=3\)
3. \(\log_{3}\frac{1}{9}=-2\)
4. \(\log_{7} 1=0\)

2. Log to exponential form

Write the base raised to the logarithm; the result is the number inside the log.

1. \(4^{3}=64\)
2. \(5^{-2}=\frac{1}{25}\)
3. \(9^{1/2}=3\)
4. \(10^{-2}=0.01\)

3. Evaluate simple logs

Ask which power of the base gives the number.

1. \(2^{4}=16\), so 4
2. \(3^{4}=81\), so 4
3. \(5^{3}=125\), so 3
4. \(10^{5}=100000\), so 5

4. Evaluate fractions and roots

1. \(2^{-3}=\frac{1}{8}\), so -3
2. \(4^{1/2}=2\), so \(\frac{1}{2}\)
3. Write \(8=2^{3}\) and \(4=2^{2}\): \(2^{3y}=2^{2}\), so \(3y=2\) and the value is \(\frac{2}{3}\)
4. \((\frac{1}{2})^{-3}=2^{3}=8\), so -3

5. Special values

1. Any base to the power 0 is 1, so 0
2. \(12^{1}=12\), so 1
3. \(\log_{b} b^{x}=x\), so 5
4. \(b^{\log_{b} x}=x\), so 11

Algebra 2 Logarithms Worksheet Answer Key: problems 6–10

6. Find the base or the value

1. \(b^{2}=49\), so \(b=7\) (the base must be positive)
2. \(b^{3}=64\), so \(b=4\)
3. \(x=3^{4}\), so \(x=81\)
4. \(x=2^{-5}\), so \(x=\frac{1}{32}\)

7. Common logarithms

1. \(10^{3}=1000\), so 3
2. \(10^{-3}=0.001\), so -3
3. \(10^{1/2}=\sqrt{10}\), so \(\frac{1}{2}\)
4. \(10^{2}=100 < 450 < 1000=10^{3}\), so \(\log 450\) is between 2 and 3 (a calculator gives \(\log 450\approx 2.653\))

8. Natural logarithms

1. 4
2. \(e^{0}=1\), so 0
3. 1
4. \(\frac{1}{e^{3}}=e^{-3}\), so -3
5. 9

9. Use a calculator

1. 1.9459
2. 1.7482
3. -0.6931 (negative because \(0.5 < 1\))

10. Read a log graph

1. The curve crosses the x-axis at \((1,0)\).
2. Domain \(x > 0\); range all real numbers.
3. The curve approaches the y-axis: \(x=0\).
4. The points \((9,2)\) and \((\frac{1}{3},-1)\) are on the graph, so \(\log_{3} 9=2\) and \(\log_{3}\frac{1}{3}=-1\).

Algebra 2 Logarithms Worksheet Answer Key: problems 11–15

11. Domain of a log function

The argument of a logarithm must be positive.

1. \(x-3 > 0\), so \(x > 3\)
2. \(5-x > 0\), so \(x < 5\)
3. \(2x+6 > 0\), so \(x > -3\): \(x > -3\)
4. \(x^{2}-9 > 0\) means \((x-3)(x+3) > 0\), so \(x < -3\) or \(x > 3\)

12. Transformations of a log graph

1. The basic graph is shifted 1 unit left, so the asymptote is \(x=-1\).
2. Domain: \(x > -1\).
3. \(g(0)=\log_{2} 1-3=-3\), so the y-intercept is \((0,-3)\).
4. \(\log_{2}(x+1)=3\), so \(x+1=8\) and the x-intercept is \((7,0)\).

13. Expand a logarithm

1. \(\log_{3} 9+\log_{3} x=\)\(2+\log_{3} x\)
2. \(3\log_{2} x-\log_{2} y\)
3. \(\log_{5} 25+\log_{5} x^{1/2}=\)\(2+\frac{1}{2}\log_{5} x\)
4. \(\ln e^{2}+4\ln x-3\ln y=\)\(2+4\ln x-3\ln y\)

14. Condense a logarithm

1. \(\log_{2} 30\)
2. \(\log 2^{3}+\log 5=\log 8+\log 5=\)\(\log 40\)
3. \(\ln 3\)
4. \(\log_{3}(x^{2}(x+1))\)
5. \(\log_{7}\frac{\sqrt{x}}{y^{3}}\)

15. Evaluate with properties

1. \(\log_{6} 36=\)2
2. \(\log_{2}\frac{96}{3}=\log_{2} 32=\)5
3. \(\log_{8} 2^{3}=\log_{8} 8=\)1
4. \(\log_{3} 36-\log_{3} 4=\log_{3} 9=\)2

Algebra 2 Logarithms Worksheet Answer Key: problems 16–20

16. Change of base

1. \(\frac{\ln 20}{\ln 3}\approx\frac{2.9957}{1.0986}\approx\)2.7268
2. \(\frac{\ln 8}{\ln 5}\approx\frac{2.0794}{1.6094}\approx\)1.2920
3. \(\frac{\ln 500}{\ln 12}\approx\frac{6.2146}{2.4849}\approx\)2.5009

17. Find an inverse function

1. Swap and solve: \(x=3^{y}\), so \(f^{-1}(x)=\log_{3} x\).
2. Swap the coordinates: \((1,0)\), \((3,1)\), \((9,2)\).
3. \(x=2\cdot 4^{y}\), so \(\frac{x}{2}=4^{y}\) and \(g^{-1}(x)=\log_{4}\frac{x}{2}\). Check: \(g(1)=8\) and \(\log_{4} 4=1\).
4. \(x=\log_{2} y+3\), so \(\log_{2} y=x-3\) and \(h^{-1}(x)=2^{x-3}\). Check: \(h(8)=6\) and \(2^{3}=8\).

18. Earthquake magnitudes

1. \(\frac{I_{7}}{I_{5}}=\frac{10^{7}}{10^{5}}=10^{2}=\)100 times as intense.
2. \(\frac{I}{I_{0}}=10^{6.4}\approx 2.51\times 10^{6}\), so about 2,510,000.

19. Acidity and pH

1. \(-\log 10^{-2}=\)2
2. \([H^{+}]=10^{-5}=\)0.00001 mol/L
3. \(\frac{10^{-2}}{10^{-5}}=10^{3}=\)1000 times
4. \(-\log(3.2\times 10^{-5})=5-\log 3.2\approx 5-0.5051=\)4.49

20. Sound levels

1. \(10\log\frac{I}{I_{0}}=60\), so \(\log\frac{I}{I_{0}}=6\) and the ratio is \(10^{6}\).
2. \(\log\frac{I}{I_{0}}=9\), so the ratio is \(10^{9}\).
3. \(\frac{10^{9}}{10^{6}}=\)1000 times .
4. \(L=10\log(2\cdot 10^{9})=10(\log 2+9)\approx 10(0.3010+9)=\)93.01 dB.

Algebra 2 Logarithms Worksheet Answer Key: problems 21–23

21. Digits of a large number

For a positive whole number N, rounding \(\log N\) down to a whole number gives one less than the number of digits of N.

\(\log 2^{50}=50\log 2\approx 50(0.30103)=15.0515\). Round down to 15 and add 1: 16 digits. Check: 2^50 = 1,125,899,906,842,624 has 16 digits.

22. Invert a growth model

\(\frac{P}{500}=2^{t/3}\), so \(\frac{t}{3}=\log_{2}\frac{P}{500}\) and \(t=3\log_{2}\frac{P}{500}\).

For \(P=4000\): \(t=3\log_{2} 8=3\cdot 3=\)9 hours. Check: \(500\cdot 2^{3}=4000\).

23. Find the error

The product rule applies to \(\log_{2}(3x)\), not to a sum inside the log.

For \(x=5\): \(\log_{2}(5+3)=\log_{2} 8=3\), but \(\log_{2} 5+\log_{2} 3\approx 2.3219+1.5850=3.9069\). The two sides are not equal.

What are the answers for the evaluation problems?

Problem 3 gives 4, 4, 3 and 5. For problem 4, the values are \(-3\), \(\frac{1}{2}\), \(\frac{2}{3}\) and \(-3\). Special values in problem 5 are 0, 1, 5 and 11. Problem 6 finds \(b=7\), then \(b=4\), \(x=81\) and \(x=\frac{1}{32}\). Natural logs in problem 8 give 4, 0, 1, \(-3\) and 9.

Which calculator and domain answers matter most?

With a calculator, \(\ln 7\approx 1.9459\), \(\log 56\approx 1.7482\) and \(\ln 0.5\approx -0.6931\). The domains in problem 11 are \(x>3\), \(x<5\) and \(x>-3\). For problem 12, the asymptote is \(x=-1\), the y-intercept is \(-3\) and the x-intercept is 7. Also, change of base gives \(\log_{3} 20\approx 2.7268\).

How do the application answers read?

The number \(2^{50}\) has 16 digits. The inverse of \(500\cdot 2^{t/3}\) is \(t=3\log_{2}\frac{P}{500}\). In the error problem, the substitution \(x=5\) shows \(\log_{2} 8=3\), yet \(\log_{2} 5+\log_{2} 3=\log_{2} 15\approx 3.91\). Therefore the claimed identity fails.

How do you turn mistakes into progress?

For each wrong answer, write the base, the exponent and the result in one line. That simple record shows whether you misread the base or misapplied a rule. Afterward, redo the problem and compare it with the key one more time.

See every chapter on the Algebra 2 math page.