Solve Exponential and Logarithmic Equations

$\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill 2{\log }_{5}x& =\hfill & {\log }_{5}81\hfill \\ \text{Use the Power Property.}\hfill & & & & & \hfill {\log }_5{x}^2& =\hfill & {\log }_{5}81\hfill \\ \text{Use the One-to-One Property, if}{\log }_{a}M={\log }_{a}N,\hfill & & & & & \hfill x^2& =\hfill & 81\hfill \\ \text{then}M=N.\hfill & & & & & \\ \text{Solve using the Square Root Property.}\hfill & & & & & \hfill x& =\hfill & \text{±}9\hfill \end{array}\hfill \\ \text{We eliminate}x=\mathrm{-9}\text{as we cannot take the logarithm}x=9,\cancel{x=\mathrm{-9}}\hfill & & & & & \\ \text{of a negative number.}\hfill & & & & & \\ \text{Check.}\hfill & & & & & \\ \begin{array}{}\\ \\ x=9\hfill & & & & & \hfill 2{\log }_{5}x& =\hfill & {\log }_{5}81\hfill \\ & & & & & \hfill 2{\log }_{5}9& \stackrel{?}{=}\hfill & {\log }_{5}81\hfill \\ & & & & & \hfill {\log }_5{9}^2& \stackrel{?}{=}\hfill & {\log }_{5}81\hfill \\ & & & & & \hfill {\log }_{5}81& =\hfill & {\log }_{5}81✓\hfill \end{array}\hfill & & & & & \end{array}$

$\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill {\log }_{3}x+{\log }_3\left(x-8\right)& =\hfill & 2\hfill \\ \text{Use the Product Property,}{\log }_{a}M+{\log }_{a}N={\log }_{a}M\cdot N.\hfill & & & & & \hfill {\log }_{3}x\left(x-8\right)& =\hfill & 2\hfill \\ \text{Rewrite in exponential form.}\hfill & & & & & \hfill 3^2& =\hfill & x\left(x-8\right)\hfill \\ \text{Simplify.}\hfill & & & & & \hfill 9& =\hfill & x^2-8x\hfill \\ \text{Subtract 9 from each side.}\hfill & & & & & \hfill 0& =\hfill & x^2-8x-9\hfill \\ \text{Factor.}\hfill & & & & & \hfill 0& =\hfill & \left(x-9\right)\left(x+1\right)\hfill \\ \text{Use the Zero-Product Property.}\hfill & & & & & \hfill x-9& =\hfill & 0,x+1=0\hfill \end{array}\hfill \\ \text{Solve each equation.}x=9,\cancel{x=\mathrm{-1}}\hfill & & & & & \\ \text{Check.}\hfill & & & & & \\ \begin{array}{}\\ \\ x=\mathrm{-1}\hfill & & & & & \hfill {\log }_{3}x+{\log }_3\left(x-8\right)& =\hfill & 2\hfill \\ & & & & & \hfill {\log }_3\left(\mathrm{-1}\right)+{\log }_3\left(\mathrm{-1}\mathrm{-8}\right)& \stackrel{?}{=}\hfill & 2\hfill \end{array}\hfill & & & & & \\ \text{We cannot take the log of a negative number.}\hfill & & & & & \\ \begin{array}{}\\ \\ x=9\hfill & & & & & \hfill {\log }_{3}x+{\log }_3\left(x-8\right)& =\hfill & 2\hfill \\ & & & & & \hfill {\log }_{3}9+{\log }_3\left(9-8\right)& \stackrel{?}{=}\hfill & 2\hfill \\ & & & & & \hfill 2+0& \stackrel{?}{=}\hfill & 2\hfill \\ & & & & & \hfill 2& =\hfill & 2✓\hfill \end{array}\hfill & & & & & \end{array}$

$\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill {\text{log}}_4\left(x+6\right)-{\text{log}}_4\left(2x+5\right)& =\hfill & \text{−}{\text{log}}_{4}x\hfill \\ \begin{array}{c}\text{Use the Quotient Property on the left side and the Power}\hfill \\ \text{Property on the right.}\hfill \end{array}\hfill & & & & & \hfill {\text{log}}_4\left(\frac{x+6}{2x+5}\right)& =\hfill & {\text{log}}_4{x}^{\mathrm{-1}}\hfill \\ \text{Rewrite}{x}^{\mathrm{-1}}=\frac{1}{x}.\hfill & & & & & \hfill {\text{log}}_4\left(\frac{x+6}{2x+5}\right)& =\hfill & {\text{log}}_4\frac{1}{x}\hfill \\ \begin{array}{c}\text{Use the One-to-One Property, if}{\text{log}}_{a}M={\text{log}}_{a}N,\hfill \\ \text{then}M=N.\hfill \end{array}\hfill & & & & & \hfill \frac{x+6}{2x+5}& =\hfill & \frac{1}{x}\hfill \\ \text{Solve the rational equation.}\hfill & & & & & \hfill x\left(x+6\right)& =\hfill & 2x+5\hfill \\ \text{Distribute.}\hfill & & & & & \hfill x^2+6x& =\hfill & 2x+5\hfill \\ \text{Write in standard form.}\hfill & & & & & \hfill x^2+4x-5& =\hfill & 0\hfill \\ \text{Factor.}\hfill & & & & & \hfill \left(x+5\right)\left(x-1\right)& =\hfill & 0\hfill \end{array}\hfill \\ \text{Use the Zero-Product Property.}x+5=0,x-1=0\hfill & & & & & \\ \text{Solve each equation.}\cancel{x=\mathrm{-5}},x=1\hfill & & & & & \\ \text{Check.}\hfill & & & & & \\ \text{We leave the check for you.}\hfill & & & & & \end{array}$

$\begin{array}{cccccccc}& & & & & \hfill 5^x& =\hfill & 11\hfill \\ \text{Since the exponential is isolated, take the logarithm of both sides.}\hfill & & & & & \hfill \text{log}{5}^x& =\hfill & \text{log}11\hfill \\ \text{Use the Power Property to get the}x\text{as a factor, not an exponent.}\hfill & & & & & \hfill x\text{log}5& =\hfill & \text{log}11\hfill \\ \text{Solve for}x.\text{Find the exact answer.}\hfill & & & & & \hfill x& =\hfill & \frac{\text{log}11}{\text{log}5}\hfill \\ \text{Approximate the answer.}\hfill & & & & & \hfill x& \approx \hfill & 1.490\hfill \\ \text{Since}{5}^1=5\text{and}{5}^2=25,\text{does it makes sense that}{5}^{1.490}\approx 11?\hfill & & & & & \end{array}$

$\begin{array}{cccccccc}& & & & & \hfill 3e^{x+2}& =\hfill & 24\hfill \\ \text{Isolate the exponential by dividing both sides by 3.}\hfill & & & & & \hfill e^{x+2}& =\hfill & 8\hfill \\ \text{Take the natural logarithm of both sides.}\hfill & & & & & \hfill \text{ln}{e}^{x+2}& =\hfill & \text{ln}8\hfill \\ \text{Use the Power Property to get the}x\text{as a factor, not an exponent.}\hfill & & & & & \hfill \left(x+2\right)\text{ln}e& =\hfill & \text{ln}8\hfill \\ \text{Use the property}\text{ln}e=1\text{to simplify.}\hfill & & & & & \hfill x+2& =\hfill & \text{ln}8\hfill \\ \text{Solve the equation. Find the exact answer.}\hfill & & & & & \hfill x& =\hfill & \text{ln}8-2\hfill \\ \text{Approximate the answer.}\hfill & & & & & \hfill x& \approx \hfill & 0.079\hfill \end{array}$

$\begin{array}{c}\begin{array}{cccccccc}& & & & & \hfill A& =\hfill & \text{\$}\mathrm{50,000}\hfill \\ & & & & & \hfill P& =\hfill & \text{\$}\mathrm{10,000}\hfill \\ \text{Identify the variables in the formula.}\hfill & & & & & \hfill r& =\hfill & ?\hfill \\ & & & & & \hfill t& =\hfill & 17\text{years}\hfill \\ & & & & & \hfill A& =\hfill & P{e}^{rt}\hfill \\ \text{Substitute the values into the formula.}\hfill & & & & & \hfill \mathrm{50,000}& =\hfill & \mathrm{10,000}{e}^{r·17}\hfill \\ \text{Solve for}r.\text{Divide each side by 10,000.}\hfill & & & & & \hfill 5& =\hfill & e^{17r}\hfill \\ \text{Take the natural log of each side.}\hfill & & & & & \hfill \text{ln}5& =\hfill & \text{ln}{e}^{17r}\hfill \\ \text{Use the Power Property.}\hfill & & & & & \hfill \text{ln}5& =\hfill & 17r\text{ln}e\hfill \\ \text{Simplify.}\hfill & & & & & \hfill \text{ln}5& =\hfill & 17r\hfill \\ \text{Divide each side by 17.}\hfill & & & & & \hfill \frac{\text{ln}5}{17}& =\hfill & r\hfill \\ \text{Approximate the answer.}\hfill & & & & & \hfill r& \approx \hfill & 0.095\hfill \\ \text{Convert to a percentage.}\hfill & & & & & \hfill r& \approx \hfill & 9.5\text{\%}\hfill \end{array}\hfill \\ \\ \text{They need the rate of growth to be approximately}9.5\text{\%}.\hfill \end{array}$

This problem requires two main steps. First we must find the unknown rate, k. Then we use that value of k to help us find the unknown number of bacteria.

$\begin{array}{cccccc}\text{Identify the variables in the formula.}\hfill & & & & & \begin{array}{ccc}\hfill A& =\hfill & 300\hfill \\ \hfill A_0& =\hfill & 100\hfill \\ \hfill k& =\hfill & ?\hfill \\ \hfill t& =\hfill & 3\text{hours}\hfill \\ \hfill A& =\hfill & A_0{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute the values in the formula.}\hfill & & & & & 300=100e^{k·3}\hfill \\ \text{Solve for}k.\text{Divide each side by 100.}\hfill & & & & & 3=e^{3k}\hfill \\ \text{Take the natural log of each side.}\hfill & & & & & \text{ln}3=\text{ln}{e}^{3k}\hfill \\ \text{Use the Power Property.}\hfill & & & & & \text{ln}3=3k\text{ln}e\hfill \\ \text{Simplify.}\hfill & & & & & \text{ln}3=3k\hfill \\ \text{Divide each side by 3.}\hfill & & & & & \frac{\text{ln}3}{3}=k\hfill \\ \text{Approximate the answer.}\hfill & & & & & k\approx 0.366\hfill \\ \\ \\ \begin{array}{c}\text{We use this rate of growth to predict the number of}\hfill \\ \text{bacteria there will be in 24 hours.}\hfill \end{array}\hfill & & & & & \begin{array}{ccc}\hfill A& =\hfill & ?\hfill \\ \hfill A_0& =\hfill & 100\hfill \\ \hfill k& =\hfill & \frac{\text{ln}3}{3}\hfill \\ \hfill t& =\hfill & 24\text{hours}\hfill \\ \hfill A& =\hfill & A_0{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute in the values.}\hfill & & & & & A=100e^{\frac{\text{ln}3}{3}·24}\hfill \\ \text{Evaluate.}\hfill & & & & & A\approx \mathrm{656,100}\hfill \\ & & & & & \text{At this rate of growth, they can expect 656,100 bacteria.}\hfill \end{array}$

This problem requires two main steps. First we must find the decay constant k. If we start with 100-mg, at the half-life there will be 50-mg remaining. We will use this information to find k. Then we use that value of k to help us find the amount of sample that will be left in 500 years.

$\begin{array}{cccccc}\text{Identify the variables in the formula.}\hfill & & & & & \begin{array}{ccc}\hfill A& =\hfill & 50\hfill \\ \hfill A_0& =\hfill & 100\hfill \\ \hfill k& =\hfill & ?\hfill \\ \hfill t& =\hfill & 1590\text{years}\hfill \\ \hfill A& =\hfill & A_0{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute the values in the formula.}\hfill & & & & & 50=100e^{k·1590}\hfill \\ \text{Solve for}k.\text{Divide each side by 100.}\hfill & & & & & 0.5=e^{1590k}\hfill \\ \text{Take the natural log of each side.}\hfill & & & & & \text{ln}0.5=\text{ln}{e}^{1590k}\hfill \\ \text{Use the Power Property.}\hfill & & & & & \text{ln}0.5=1590k\text{ln}e\hfill \\ \text{Simplify.}\hfill & & & & & \text{ln}0.5=1590k\hfill \\ \text{Divide each side by 1590.}\hfill & & & & & \frac{\text{ln}0.5}{1590}=k\text{exact answer}\hfill \\ \\ \\ \begin{array}{c}\text{We use this rate of growth to predict the amount}\hfill \\ \text{that will be left in 500 years.}\hfill \end{array}\hfill & & & & & \begin{array}{ccc}\hfill A& =\hfill & ?\hfill \\ \hfill A_0& =\hfill & 100\hfill \\ \hfill k& =\hfill & \frac{\text{ln}0.5}{1590}\hfill \\ \hfill t& =\hfill & 500\text{years}\hfill \\ \hfill A& =\hfill & A_0{e}^{kt}\hfill \end{array}\hfill \\ \text{Substitute in the values.}\hfill & & & & & A=100e^{\frac{\text{ln}0.5}{1590}·500}\hfill \\ \text{Evaluate.}\hfill & & & & & A\approx 80.4\text{mg}\hfill \\ & & & & & \begin{array}{c}\text{In 500 years there would be}\hfill \\ \text{approximately 80.4 mg remaining.}\hfill \end{array}\hfill \end{array}$