6.2E: Exercises
- Page ID
- 120174
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Exercises
In Exercises 1 - 15, expand the given logarithm and simplify. Assume when necessary that all quantities represent positive real numbers.
- \(\ln(x^{3}y^{2})\)
- \(\log_{2}\left(\dfrac{128}{x^{2} + 4}\right)\)
- \(\log_{5}\left(\dfrac{z}{25}\right)^{3}\)
- \(\log(1.23 \times 10^{37})\)
- \(\ln\left(\dfrac{\sqrt{z}}{xy}\right)\)
- \(\log_{5} \left(x^2 - 25 \right)\)
- \(\log_{\sqrt{2}} \left(4x^3\right)\)
- \(\log_{\frac{1}{3}}(9x(y^{3} - 8))\)
- \(\log\left(1000x^3y^5\right)\)
- \(\log_{3} \left(\dfrac{x^2}{81y^4}\right)\)
- \(\ln\left(\sqrt[4]{\dfrac{xy}{ez}}\right)\)
- \(\log_{6} \left(\dfrac{216}{x^3y}\right)^4\)
- \(\log\left(\dfrac{100x\sqrt{y}}{\sqrt[3]{10}}\right)\)
- \(\log_{\frac{1}{2}}\left(\dfrac{4\sqrt[3]{x^2}}{y\sqrt{z}}\right)\)
- \(\ln \left(\dfrac{\sqrt[3]{x}}{10 \sqrt{yz}}\right)\)
In Exercises 16 - 29, use the properties of logarithms to write the expression as a single logarithm.
- \(4\ln(x) + 2\ln(y)\)
- \(\log_{2}(x) + \log_{2}(y) - \log_{2}(z)\)
- \(\log_{3}(x) - 2 \log_{3}(y)\)
- \(\frac{1}{2}\log_{3}(x) - 2\log_{3}(y) - \log_{3}(z)\)
- \(2 \ln(x) -3 \ln(y) - 4\ln(z)\)
- \(\log(x) - \frac{1}{3} \log(z) + \frac{1}{2} \log(y)\)
- \(-\frac{1}{3} \ln(x) - \frac{1}{3}\ln(y) + \frac{1}{3} \ln(z)\)
- \(\log_{5}(x) - 3\)
- \(3 - \log(x)\)
- \(\log_{7}(x) + \log_{7}(x - 3) - 2\)
- \(\ln(x) + \frac{1}{2}\)
- \(\log_{2}(x) + \log_{4}(x)\)
- \(\log_{2}(x) + \log_{4}(x-1)\)
- \(\log_{2}(x) + \log_{\frac{1}{2}}(x - 1)\)
In Exercises 30 - 33, use the appropriate change of base formula to convert the given expression to an expression with the indicated base.
- \(7^{x - 1}\) to base \(e\)
- \(\log_{3}(x + 2)\) to base 10
- \(\left(\dfrac{2}{3}\right)^{x}\) to base \(e\)
- \(\log(x^{2} + 1)\) to base \(e\)
In Exercises 34 - 39, use the appropriate change of base formula to approximate the logarithm.
- \(\log_{3}(12)\)
- \(\log_{5}(80)\)
- \(\log_{6}(72)\)
- \(\log_{4}\left(\dfrac{1}{10}\right)\)
- \(\log_{\frac{3}{5}}(1000)\)
- \(\log_{\frac{2}{3}}(50)\)
- Compare and contrast the graphs of \(y = \ln(x^{2})\) and \(y = 2\ln(x)\).
- Prove the Quotient Rule and Power Rule for Logarithms.
- Give numerical examples to show that, in general,
- \(\log_{b}(x + y) \neq \log_{b}(x) + \log_{b}(y)\)
- \(\log_{b}(x - y) \neq \log_{b}(x) - \log_{b}(y)\)
- \(\log_{b}\left(\dfrac{x}{y}\right) \neq \dfrac{\log_{b}(x)}{\log_{b}(y)}\)
- The Henderson-Hasselbalch Equation: Suppose \(HA\) represents a weak acid. Then we have a reversible chemical reaction \[HA \rightleftharpoons H^{+} + A^{-}. \nonumber\] The acid disassociation constant, \(K_{a}\), is given by \[K_{\alpha} = \frac{[H^{+}][A^{-}]}{[HA]} = [H^{+}]\frac{[A^{-}]}{[HA]},\nonumber\] where the square brackets denote the concentrations just as they did in Exercise 77 in Section 6.1. The symbol p\(K_{a}\) is defined similarly to pH in that p\(K_{a} = -\log(K_{a})\). Using the definition of pH from Exercise 77 and the properties of logarithms, derive the Henderson-Hasselbalch Equation which states \[\mbox{pH} = \mbox{p}K_{a} + \log\dfrac{[A^{-}]}{[HA]}\nonumber\nonumber\]
- Research the history of logarithms including the origin of the word ‘logarithm’ itself. Why is the abbreviation of natural log ‘ln’ and not ‘nl’?
- There is a scene in the movie "Apollo 13" in which several people at Mission Control use slide rules to verify a computation. Was that scene accurate? Look for other pop culture references to logarithms and slide rules.
Answers
- \(3\ln(x) + 2\ln(y)\)
- \(7 - \log_{2}(x^{2} + 4)\)
- \(3\log_{5}(z) - 6\)
- \(\log(1.23) + 37\)
- \(\frac{1}{2}\ln(z) - \ln(x) - \ln(y)\)
- \(\log_{5}(x-5) + \log_{5}(x+5)\)
- \(3\log_{\sqrt{2}}(x) + 4\)
- \(-2 + \log_{\frac{1}{3}}(x) + \log_{\frac{1}{3}}(y - 2) + \log_{\frac{1}{3}}(y^{2} + 2y + 4)\)
- \(3 + 3\log(x) + 5 \log(y)\)
- \(2\log_{3}(x) - 4 - 4\log_{3}(y)\)
- \(\frac{1}{4} \ln(x) + \frac{1}{4} \ln(y) - \frac{1}{4} - \frac{1}{4} \ln(z)\)
- \(12-12\log_{6}(x) - 4\log_{6}(y)\)
- \(\frac{5}{3}+\log(x)+\frac{1}{2}\log(y)\)
- \(-2+\frac{2}{3}\log_{\frac{1}{2}}(x)-\log_{\frac{1}{2}}(y)-\frac{1}{2}\log_{\frac{1}{2}}(z)\)
- \(\frac{1}{3} \ln(x) - \ln(10) - \frac{1}{2}\ln(y)-\frac{1}{2}\ln(z)\)
- \(\ln(x^{4}y^{2})\)
- \(\log_{2}\left(\frac{xy}{z}\right)\)
- \(\log_{3} \left( \frac{x}{y^2} \right)\)
- \(\log_{3}\left(\frac{\sqrt{x}}{y^{2}z}\right)\)
- \(\ln\left( \frac{x^2}{y^3z^4} \right)\)
- \(\log\left(\frac{x \sqrt{y}}{\sqrt[3]{z}} \right)\)
- \(\ln\left(\sqrt[3]{\frac{z}{xy}} \right)\)
- \(\log_{5}\left(\frac{x}{125}\right)\)
- \(\log\left(\frac{1000}{x}\right)\)
- \(\log_{7}\left(\frac{x(x - 3)}{49}\right)\)
- \(\ln \left(x \sqrt{e} \right)\)
- \(\log_{2}\left(x^{3/2}\right)\)
- \(\log_{2}\left(x \sqrt{x-1}\right)\)
- \(\vphantom{\frac{\log(x + 2)}{\log(3)}}\log_{2}\left(\frac{x}{x - 1}\right)\)
- \(\vphantom{\frac{\log(x + 2)}{\log(3)}}7^{x - 1} = e^{(x - 1)\ln(7)}\)
- \(\log_{3}(x + 2) = \frac{\log(x + 2)}{\log(3)}\)
- \(\left(\frac{2}{3}\right)^{x} = e^{x\ln(\frac{2}{3})}\)
- \(\log(x^{2} + 1) = \frac{\ln(x^{2} + 1)}{\ln(10)}\)
- \(\log_{3}(12) \approx 2.26186\)
- \(\log_{5}(80) \approx 2.72271\)
- \(\log_{6}(72) \approx 2.38685\)
- \(\log_{4}\left(\frac{1}{10}\right) \approx -1.66096\)
- \(\log_{\frac{3}{5}}(1000) \approx -13.52273\)
- \(\log_{\frac{2}{3}}(50) \approx -9.64824\)