Logarithmic Functions: practice solutions, Grade 11 – download the PDF

1 Logarithmic form ★★★

Use \(b^y=x \iff \log_b x=y\): the base stays the base, and the exponent becomes the value of the logarithm.

1. \(\log_2 64=6\)
2. \(\log_{10}0.01=-2\)
3. \(\log_7 49=2\)
4. \(\log_5 1=0\)
5. \(\log_9 3=\tfrac12\)

2 Exponential form ★★★

1. \(3^4=81\)
2. \(5^3=125\)
3. \(2^{-3}=\tfrac18\)
4. \(16^{1/2}=4\)
5. \(10^3=1000\) (the base of \(\log\) is 10)

3 Evaluating logarithms ★★★

1. \(2^4=16\), so the value is \(4\).
2. \(5^{-2}=\tfrac1{25}\), so \(-2\).
3. \(7^1=7\), so \(1\).
4. \(9^0=1\), so \(0\).
5. \(4^{1/2}=2\), so \(\tfrac12\).
6. \(\left(\tfrac12\right)^{-3}=8\), so \(-3\).

4 True or false? ★★★

1. False. \(\log_2 0\) is undefined: no power of 2 equals 0. (Note \(\log_2 1=0\).)
2. True. A power of 3 is never negative.
3. True. \(\log_b b^x=x\).
4. True. \(10^2=100\).
5. False. \(\log_2 8+\log_2 4=3+2=5=\log_2 32\), and \(\log_2 12\approx 3.58\).

5 Domain of a logarithm ★★★

The expression inside the logarithm must be positive.

1. \(x-4\gt 0\), so \(x\gt 4\).
2. \(2x+10\gt 0\), so \(x\gt -5\).
3. \(8-x\gt 0\), so \(x\lt 8\).

6 Splitting products ★★★

1. \(\log_2 8+\log_2 x=3+\log_2 x\).
2. \(\log 1000+\log y^2=3+2\log y\).
3. \(\log_5 x^3-\log_5 25=3\log_5 x-2\).

7 Natural logarithm values ★★★

1. \(\ln e^4=4\).
2. \(e^{\ln 9}=9\).
3. \(\ln 1=0\).
4. \(\ln e^{-1}=-1\).
5. \(\ln e^{1/2}=\tfrac12\).

8 Condensing logarithms ★★★

1. \(\log_3(4\cdot5)=\log_3 20\).
2. \(\log_2\dfrac{48}{3}=\log_2 16=4\).
3. \(\log 25+\log 4=\log 100=2\).
4. \(\ln 81^{1/2}-\ln 3=\ln 9-\ln 3=\ln 3\).

9 Change of base ★★★

1. \(\dfrac{\ln 20}{\ln 3}\approx\dfrac{2.9957}{1.0986}\approx 2.727\).
2. \(\dfrac{\ln 50}{\ln 7}\approx\dfrac{3.9120}{1.9459}\approx 2.010\).
3. \(\dfrac{\ln 10}{\ln 0.5}\approx\dfrac{2.3026}{-0.6931}\approx -3.322\). The result is negative because the base is less than 1 and 10 is greater than 1.

10 Same base ★★★

1. \(5^x=5^3\), so \(x=3\).
2. \(4^{x+1}=4^3\), so \(x+1=3\) and \(x=2\).
3. \(\tfrac1{32}=2^{-5}\), so \(3x-1=-5\), \(3x=-4\), \(x=-\tfrac43\).
4. \(9^x=3^{2x}\) and \(27=3^3\), so \(2x=3\) and \(x=\tfrac32\).

11 Simple logarithmic equations ★★★

1. \(x+3=2^5=32\), so \(x=29\). Check: \(x+3=32\gt 0\).
2. \(2x-1=25\), so \(x=13\).
3. \(x=10^{-2}=0.01\).
4. \(x=e^3\approx 20.09\).

12 Reading a logarithmic graph ★★★

1. We need \(x+5\gt 0\): the domain is \(x\gt -5\) and the asymptote is \(x=-5\) (the parent graph shifted 5 units left and 3 units down).
2. \(f(x)=0 \iff \log_2(x+5)=3 \iff x+5=8\), so \(x=3\). The x-intercept is \((3,0)\).
3. \(f(0)=\log_2 5-3=\dfrac{\ln 5}{\ln 2}-3\approx 2.322-3\approx -0.678\).
4. Three convenient points: \((-4,-3)\), \((-1,-1)\), \((3,0)\). The curve rises slowly and hugs the asymptote on the left.

13 Using a graph and logs ★★★

1. The curve meets the line a little before \(x=3\) (since \(3^3=27\gt 20\)) and after \(x=2\) (since \(3^2=9\lt 20\)): about 2.7.
2. \(3^x=20\) gives \(x=\log_3 20=\dfrac{\ln 20}{\ln 3}\approx 2.727\), which agrees with the graph.

14 Growing bacteria ★★★

1. \(500\cdot 2^{t/3}=8000\) gives \(2^{t/3}=16=2^4\), so \(t/3=4\) and \(t=12\) hours.
2. \(2^{t/3}=20\), so \(\dfrac t3=\log_2 20=\dfrac{\ln 20}{\ln 2}\approx 4.322\) and \(t\approx 12.97\), about 13.0 hours.

15 A product of logs ★★★

Condense: \(\log_2\big(x(x-6)\big)=4\), so \(x^2-6x=16\), i.e. \(x^2-6x-16=0\), i.e. \((x-8)(x+2)=0\).

Candidates: \(x=8\) and \(x=-2\). For \(x=-2\), \(\log_2 x\) and \(\log_2(x-6)\) would involve negative numbers, which is impossible, so \(-2\) is extraneous. Check \(x=8\): \(\log_2 8+\log_2 2=3+1=4\). The solution is \(x=8\).

16 A quotient of logs ★★★

Condense: \(\log\dfrac{x+3}{x-1}=1\), so \(\dfrac{x+3}{x-1}=10\).

Then \(x+3=10x-10\), so \(13=9x\) and \(x=\dfrac{13}{9}\approx 1.444\).

Check: \(x\gt 1\), so both \(x+3\) and \(x-1\) are positive. The solution is \(x=\tfrac{13}{9}\).

17 Saving for a goal ★★★

1. \((1.036)^t=1.5\), so \(t=\dfrac{\ln 1.5}{\ln 1.036}\approx\dfrac{0.4055}{0.03537}\approx 11.46\) years.
2. \(e^{0.036t}=1.5\), so \(t=\dfrac{\ln 1.5}{0.036}\approx 11.26\) years.

Continuous compounding is a little faster, by about 0.2 year (roughly 2 months).

18 Radioactive decay ★★★

1. \(\left(\tfrac12\right)^{t/12}=\tfrac1{16}=\left(\tfrac12\right)^4\), so \(t=48\) days.
2. \(\left(\tfrac12\right)^{t/12}=\tfrac14\), so \(t/12=2\) and \(t=24\) days.
3. \(\left(\tfrac12\right)^{t/12}=\tfrac38\), so \(\dfrac t{12}=\dfrac{\ln(3/8)}{\ln(1/2)}\approx 1.415\) and \(t\approx 17.0\) days.

19 A hidden quadratic ★★★

Let \(u=e^x\). The equation becomes \(u^2-5u+6=0\), i.e. \((u-2)(u-3)=0\), so \(u=2\) or \(u=3\).

Then \(e^x=2\) gives \(x=\ln 2\approx 0.693\), and \(e^x=3\) gives \(x=\ln 3\approx 1.099\). Both are valid because \(e^x\) is always positive.

20 Finding an inverse ★★★

1. Write \(y=2\cdot 3^x+1\) and swap: \(x=2\cdot 3^y+1\). Then \(3^y=\dfrac{x-1}{2}\), so \(f^{-1}(x)=\log_3\dfrac{x-1}{2}\).
2. We need \(\dfrac{x-1}{2}\gt 0\): the domain is \(x\gt 1\) (the range of \(f\)).
3. \(f^{-1}(55)=\log_3 27=3\). Check: \(f(3)=2\cdot 27+1=55\).

21 Reciprocal logarithms ★★★

1. With natural logs: \(\log_a b\cdot\log_b a=\dfrac{\ln b}{\ln a}\cdot\dfrac{\ln a}{\ln b}=1\).
2. \(\log_2 9=2\log_2 3\) and \(\log_3 8=3\log_3 2\). Since \(\log_2 3\cdot\log_3 2=1\), the product is \(2\cdot 3\cdot 1=6\).