<< Chapter < Page | Chapter >> Page > |
${\mathrm{log}}_{y}\left(137\right)=x$
$\text{log}(v)=t$
For the following exercises, rewrite each equation in logarithmic form.
${4}^{x}=y$
${m}^{-7}=n$
${x}^{-\text{\hspace{0.17em}}\frac{10}{13}}=y$
${\left(\frac{7}{5}\right)}^{m}=n$
${y}^{x}=\frac{39}{100}$
${\mathrm{log}}_{y}\left(\frac{39}{100}\right)=x$
${10}^{a}=b$
For the following exercises, solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ by converting the logarithmic equation to exponential form.
${\text{log}}_{3}(x)=2$
${\text{log}}_{5}(x)=2$
${\text{log}}_{2}(x)=6$
${\text{log}}_{18}(x)=2$
${\mathrm{log}}_{6}\left(x\right)=-3$
$x={6}^{-3}=\frac{1}{216}$
$\text{log}(x)=3$
For the following exercises, use the definition of common and natural logarithms to simplify.
$\text{log}({100}^{8})$
$2\text{log}(.0001)$
$\mathrm{ln}\left({e}^{-5.03}\right)$
For the following exercises, evaluate the base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ logarithmic expression without using a calculator.
${\text{log}}_{3}\left(\frac{1}{27}\right)$
${\text{log}}_{2}\left(\frac{1}{8}\right)+4$
For the following exercises, evaluate the common logarithmic expression without using a calculator.
$\text{log}(10,000)$
$\text{log}(1)+7$
For the following exercises, evaluate the natural logarithmic expression without using a calculator.
$\text{ln}({e}^{\frac{1}{3}})$
$\text{ln}({e}^{-0.225})-3$
For the following exercises, evaluate each expression using a calculator. Round to the nearest thousandth.
$\text{log}(0.04)$
$\text{ln}\left(\frac{4}{5}\right)$
$\text{ln}(\sqrt{2})$
Is $\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$ in the domain of the function $\text{\hspace{0.17em}}f(x)=\mathrm{log}(x)?\text{\hspace{0.17em}}$ If so, what is the value of the function when $\text{\hspace{0.17em}}x=0?\text{\hspace{0.17em}}$ Verify the result.
No, the function has no defined value for $\text{\hspace{0.17em}}x=0.\text{\hspace{0.17em}}$ To verify, suppose $\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$ is in the domain of the function $\text{\hspace{0.17em}}f(x)=\mathrm{log}(x).\text{\hspace{0.17em}}$ Then there is some number $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}n=\mathrm{log}(0).\text{\hspace{0.17em}}$ Rewriting as an exponential equation gives: $\text{\hspace{0.17em}}{10}^{n}=0,$ which is impossible since no such real number $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ exists. Therefore, $\text{\hspace{0.17em}}x=0\text{\hspace{0.17em}}$ is not the domain of the function $\text{\hspace{0.17em}}f(x)=\mathrm{log}(x).$
Is $\text{\hspace{0.17em}}f(x)=0\text{\hspace{0.17em}}$ in the range of the function $\text{\hspace{0.17em}}f(x)=\mathrm{log}(x)?\text{\hspace{0.17em}}$ If so, for what value of $\text{\hspace{0.17em}}x?\text{\hspace{0.17em}}$ Verify the result.
Is there a number $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}\mathrm{ln}x=2?\text{\hspace{0.17em}}$ If so, what is that number? Verify the result.
Yes. Suppose there exists a real number $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}\mathrm{ln}x=2.\text{\hspace{0.17em}}$ Rewriting as an exponential equation gives $\text{\hspace{0.17em}}x={e}^{2},$ which is a real number. To verify, let $\text{\hspace{0.17em}}x={e}^{2}.\text{\hspace{0.17em}}$ Then, by definition, $\text{\hspace{0.17em}}\mathrm{ln}\left(x\right)=\mathrm{ln}\left({e}^{2}\right)=2.$
Is the following true: $\text{\hspace{0.17em}}\frac{{\mathrm{log}}_{3}(27)}{{\mathrm{log}}_{4}\left(\frac{1}{64}\right)}=\mathrm{-1}?\text{\hspace{0.17em}}$ Verify the result.
Is the following true: $\text{\hspace{0.17em}}\frac{\mathrm{ln}\left({e}^{1.725}\right)}{\mathrm{ln}\left(1\right)}=1.725?\text{\hspace{0.17em}}$ Verify the result.
No; $\text{\hspace{0.17em}}\mathrm{ln}\left(1\right)=0,$ so $\text{\hspace{0.17em}}\frac{\mathrm{ln}\left({e}^{1.725}\right)}{\mathrm{ln}\left(1\right)}\text{\hspace{0.17em}}$ is undefined.
The exposure index $\text{\hspace{0.17em}}EI\text{\hspace{0.17em}}$ for a 35 millimeter camera is a measurement of the amount of light that hits the film. It is determined by the equation $\text{\hspace{0.17em}}EI={\mathrm{log}}_{2}\left(\frac{{f}^{2}}{t}\right),$ where $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is the “f-stop” setting on the camera, and $t$ is the exposure time in seconds. Suppose the f-stop setting is $\text{\hspace{0.17em}}8\text{\hspace{0.17em}}$ and the desired exposure time is $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ seconds. What will the resulting exposure index be?
Refer to the previous exercise. Suppose the light meter on a camera indicates an $\text{\hspace{0.17em}}EI\text{\hspace{0.17em}}$ of $\text{\hspace{0.17em}}-2,$ and the desired exposure time is 16 seconds. What should the f-stop setting be?
$2$
The intensity levels I of two earthquakes measured on a seismograph can be compared by the formula $\text{\hspace{0.17em}}\mathrm{log}\frac{{I}_{1}}{{I}_{2}}={M}_{1}-{M}_{2}\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}M\text{\hspace{0.17em}}$ is the magnitude given by the Richter Scale. In August 2009, an earthquake of magnitude 6.1 hit Honshu, Japan. In March 2011, that same region experienced yet another, more devastating earthquake, this time with a magnitude of 9.0. http://earthquake.usgs.gov/earthquakes/world/historical.php. Accessed 3/4/2014. How many times greater was the intensity of the 2011 earthquake? Round to the nearest whole number.
Notification Switch
Would you like to follow the 'Precalculus' conversation and receive update notifications?