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Given a logarithm with the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}M,$ use the change-of-base formula to rewrite it as a quotient of logs with any positive base $\text{\hspace{0.17em}}n,$ where $\text{\hspace{0.17em}}n\ne 1.$
Change $\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\text{\hspace{0.17em}}$ to a quotient of natural logarithms.
Because we will be expressing $\text{\hspace{0.17em}}{\mathrm{log}}_{5}3\text{\hspace{0.17em}}$ as a quotient of natural logarithms, the new base, $\text{\hspace{0.17em}}n=e.$
We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.
Change $\text{\hspace{0.17em}}{\mathrm{log}}_{0.5}8\text{\hspace{0.17em}}$ to a quotient of natural logarithms.
$\frac{\mathrm{ln}8}{\mathrm{ln}0.5}$
Can we change common logarithms to natural logarithms?
Yes. Remember that $\text{\hspace{0.17em}}\mathrm{log}9\text{\hspace{0.17em}}$ means $\text{\hspace{0.17em}}{\text{log}}_{\text{10}}\text{9}.$ So, $\text{\hspace{0.17em}}\mathrm{log}9=\frac{\mathrm{ln}9}{\mathrm{ln}10}.$
Evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{2}(10)\text{\hspace{0.17em}}$ using the change-of-base formula with a calculator.
According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base $\text{\hspace{0.17em}}e.$
Evaluate $\text{\hspace{0.17em}}{\mathrm{log}}_{5}(100)\text{\hspace{0.17em}}$ using the change-of-base formula.
$\frac{\mathrm{ln}100}{\mathrm{ln}5}\approx \frac{4.6051}{1.6094}=2.861$
Access these online resources for additional instruction and practice with laws of logarithms.
The Product Rule for Logarithms | ${\mathrm{log}}_{b}(MN)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)$ |
The Quotient Rule for Logarithms | ${\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N$ |
The Power Rule for Logarithms | ${\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M$ |
The Change-of-Base Formula | ${\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\text{}n0,n\ne 1,b\ne 1$ |
How does the power rule for logarithms help when solving logarithms with the form $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(\sqrt[n]{x}\right)?$
Any root expression can be rewritten as an expression with a rational exponent so that the power rule can be applied, making the logarithm easier to calculate. Thus, $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left({x}^{\frac{1}{n}}\right)=\frac{1}{n}{\mathrm{log}}_{b}(x).$
What does the change-of-base formula do? Why is it useful when using a calculator?
For the following exercises, expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.
${\mathrm{log}}_{b}\left(7x\cdot 2y\right)$
${\mathrm{log}}_{b}\left(2\right)+{\mathrm{log}}_{b}\left(7\right)+{\mathrm{log}}_{b}\left(x\right)+{\mathrm{log}}_{b}\left(y\right)$
$\mathrm{ln}\left(3ab\cdot 5c\right)$
${\mathrm{log}}_{b}\left(\frac{13}{17}\right)$
${\mathrm{log}}_{b}\left(13\right)-{\mathrm{log}}_{b}\left(17\right)$
${\mathrm{log}}_{4}\left(\frac{\text{}\frac{x}{z}\text{}}{w}\right)$
$\mathrm{ln}\left(\frac{1}{{4}^{k}}\right)$
$-k\mathrm{ln}(4)$
${\mathrm{log}}_{2}\left({y}^{x}\right)$
For the following exercises, condense to a single logarithm if possible.
$\mathrm{ln}\left(7\right)+\mathrm{ln}\left(x\right)+\mathrm{ln}\left(y\right)$
$\mathrm{ln}\left(7xy\right)$
${\mathrm{log}}_{3}(2)+{\mathrm{log}}_{3}(a)+{\mathrm{log}}_{3}(11)+{\mathrm{log}}_{3}(b)$
${\mathrm{log}}_{b}(28)-{\mathrm{log}}_{b}(7)$
${\mathrm{log}}_{b}(4)$
$\mathrm{ln}\left(a\right)-\mathrm{ln}\left(d\right)-\mathrm{ln}\left(c\right)$
$-{\mathrm{log}}_{b}\left(\frac{1}{7}\right)$
${\text{log}}_{b}\left(7\right)$
$\frac{1}{3}\mathrm{ln}\left(8\right)$
For the following exercises, use the properties of logarithms to expand each logarithm as much as possible. Rewrite each expression as a sum, difference, or product of logs.
$\mathrm{log}\left(\frac{{x}^{15}{y}^{13}}{{z}^{19}}\right)$
$15\mathrm{log}(x)+13\mathrm{log}(y)-19\mathrm{log}(z)$
$\mathrm{ln}\left(\frac{{a}^{\mathrm{-2}}}{{b}^{\mathrm{-4}}{c}^{5}}\right)$
$\mathrm{log}\left(\sqrt{{x}^{3}{y}^{-4}}\right)$
$\frac{3}{2}\mathrm{log}(x)-2\mathrm{log}(y)$
$\mathrm{ln}\left(y\sqrt{\frac{y}{1-y}}\right)$
$\mathrm{log}\left({x}^{2}{y}^{3}\sqrt[3]{{x}^{2}{y}^{5}}\right)$
$\frac{8}{3}\mathrm{log}(x)+\frac{14}{3}\mathrm{log}(y)$
For the following exercises, condense each expression to a single logarithm using the properties of logarithms.
$\mathrm{log}\left(2{x}^{4}\right)+\mathrm{log}\left(3{x}^{5}\right)$
$\mathrm{ln}(6{x}^{9})-\mathrm{ln}(3{x}^{2})$
$\mathrm{ln}(2{x}^{7})$
$2\mathrm{log}(x)+3\mathrm{log}(x+1)$
$\mathrm{log}(x)-\frac{1}{2}\mathrm{log}(y)+3\mathrm{log}(z)$
$\mathrm{log}\left(\frac{x{z}^{3}}{\sqrt{y}}\right)$
$4{\mathrm{log}}_{7}\left(c\right)+\frac{{\mathrm{log}}_{7}\left(a\right)}{3}+\frac{{\mathrm{log}}_{7}\left(b\right)}{3}$
For the following exercises, rewrite each expression as an equivalent ratio of logs using the indicated base.
${\mathrm{log}}_{7}\left(15\right)\text{\hspace{0.17em}}$ to base $\text{\hspace{0.17em}}e$
${\mathrm{log}}_{7}\left(15\right)=\frac{\mathrm{ln}\left(15\right)}{\mathrm{ln}\left(7\right)}$
${\mathrm{log}}_{14}\left(55.875\right)\text{\hspace{0.17em}}$ to base $\text{\hspace{0.17em}}10$
For the following exercises, suppose $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(6\right)=a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{\mathrm{log}}_{5}\left(11\right)=b.\text{\hspace{0.17em}}$ Use the change-of-base formula along with properties of logarithms to rewrite each expression in terms of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b.\text{\hspace{0.17em}}$ Show the steps for solving.
${\mathrm{log}}_{11}\left(5\right)$
${\mathrm{log}}_{11}\left(5\right)=\frac{{\mathrm{log}}_{5}\left(5\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{1}{b}$
${\mathrm{log}}_{6}\left(55\right)$
${\mathrm{log}}_{11}\left(\frac{6}{11}\right)$
${\mathrm{log}}_{11}\left(\frac{6}{11}\right)=\frac{{\mathrm{log}}_{5}\left(\frac{6}{11}\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{{\mathrm{log}}_{5}\left(6\right)-{\mathrm{log}}_{5}\left(11\right)}{{\mathrm{log}}_{5}\left(11\right)}=\frac{a-b}{b}=\frac{a}{b}-1$
For the following exercises, use properties of logarithms to evaluate without using a calculator.
${\mathrm{log}}_{3}\left(\frac{1}{9}\right)-3{\mathrm{log}}_{3}\left(3\right)$
$6{\mathrm{log}}_{8}\left(2\right)+\frac{{\mathrm{log}}_{8}\left(64\right)}{3{\mathrm{log}}_{8}\left(4\right)}$
$3$
$2{\mathrm{log}}_{9}\left(3\right)-4{\mathrm{log}}_{9}\left(3\right)+{\mathrm{log}}_{9}\left(\frac{1}{729}\right)$
For the following exercises, use the change-of-base formula to evaluate each expression as a quotient of natural logs. Use a calculator to approximate each to five decimal places.
${\mathrm{log}}_{8}\left(65\right)$
${\mathrm{log}}_{4}\left(\frac{15}{2}\right)$
Use the product rule for logarithms to find all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values such that $\text{\hspace{0.17em}}{\mathrm{log}}_{12}\left(2x+6\right)+{\mathrm{log}}_{12}\left(x+2\right)=2.\text{\hspace{0.17em}}$ Show the steps for solving.
Use the quotient rule for logarithms to find all $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values such that $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)=1.\text{\hspace{0.17em}}$ Show the steps for solving.
$x=4;\text{\hspace{0.17em}}$ By the quotient rule: ${\mathrm{log}}_{6}\left(x+2\right)-{\mathrm{log}}_{6}\left(x-3\right)={\mathrm{log}}_{6}\left(\frac{x+2}{x-3}\right)=1.$
Rewriting as an exponential equation and solving for $\text{\hspace{0.17em}}x:$
$$\begin{array}{ll}{6}^{1}\hfill & =\frac{x+2}{x-3}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2}{x-3}-6\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2}{x-3}-\frac{6\left(x-3\right)}{\left(x-3\right)}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x+2-6x+18}{x-3}\hfill \\ \text{\hspace{0.17em}}0\hfill & =\frac{x-4}{x-3}\hfill \\ \text{}\text{\hspace{0.17em}}x\hfill & =4\hfill \end{array}$$
Checking, we find that $\text{\hspace{0.17em}}{\mathrm{log}}_{6}\left(4+2\right)-{\mathrm{log}}_{6}\left(4-3\right)={\mathrm{log}}_{6}\left(6\right)-{\mathrm{log}}_{6}\left(1\right)\text{\hspace{0.17em}}$ is defined, so $\text{\hspace{0.17em}}x=4.$
Can the power property of logarithms be derived from the power property of exponents using the equation $\text{\hspace{0.17em}}{b}^{x}=m?\text{\hspace{0.17em}}$ If not, explain why. If so, show the derivation.
Prove that $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(n\right)=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}\text{\hspace{0.17em}}$ for any positive integers $\text{\hspace{0.17em}}b>1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n>1.$
Let $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ be positive integers greater than $\text{\hspace{0.17em}}1.\text{\hspace{0.17em}}$ Then, by the change-of-base formula, $\text{\hspace{0.17em}}{\mathrm{log}}_{b}\left(n\right)=\frac{{\mathrm{log}}_{n}\left(n\right)}{{\mathrm{log}}_{n}\left(b\right)}=\frac{1}{{\mathrm{log}}_{n}\left(b\right)}.$
Does $\text{\hspace{0.17em}}{\mathrm{log}}_{81}\left(2401\right)={\mathrm{log}}_{3}\left(7\right)?\text{\hspace{0.17em}}$ Verify the claim algebraically.
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