5.4 Solve Exponential and Logarithmic Equations

Step 1: Use the Power Property.

[latex]\log_5x^2=\log_581[/latex]

Step 2: Use the One-to-One Property, if [latex]{\mathrm{log}}_{a}M={\mathrm{log}}_{a}N[/latex] then [latex]M=N[/latex].

[latex]x^2=81[/latex]

Step 3: Solve using the Square Root Property.

[latex]x=\pm9[/latex]

Step 4: We eliminate [latex]x=-9[/latex] as e cannot take the logarithm of a negative number. 

[latex]x=9,\;\overline{)x=-9}[/latex]

Step 5: Check.

[latex]\begin{eqnarray*}x&=&9\\2\log_5x&=&\log_581\\2\log_59&\overset?=&\log_581\\\log_59^2&\overset?=&\log_581\\\log_581&=&\log_581\;\checkmark\end{eqnarray*}[/latex]

Step 1: Use the Product Property, [latex]{\mathrm{log}}_{a}M+{\mathrm{log}}_{a}N={\mathrm{log}}_{a}M\cdot N[/latex].

[latex]{\mathrm{log}}_{3}x\left(x-8\right)=2[/latex]

Step 2: Rewrite in exponential form. 

[latex]3^2\;=\;x(x-8)[/latex]

Step 3: Simplify.

[latex]9=x^2-8x[/latex]

Step 4: Subtract 9 from each side. 

[latex]0=x^2-8x-9[/latex]

Step 5: Factor. 

[latex]0=(x-9)(x+1)[/latex]

Step 6: Use the Zero-Product Property.

[latex]x-9=0,\;x+1=0[/latex]

Step 7: Solve each equation. 

[latex]x=9,\;\overline{)x=-1}[/latex]

Step 8: Check. 

[latex]\begin{eqnarray*}x&=&-1\\\log_3x+\log_3(x-8)&=&2\\\log_3({\color{red}{-}}{\color{red}{1}})+\log_3({\color{red}{-}}{\color{red}{1}}{\color{red}{-}}{\color{red}{8}})&\overset?=&2\;\text{We cannot take the log of a }{\color{red}{\text{negative}}}\text{ number.}\\\\\\x&=&{\color{blue}{9}}\\\log_3x+\log_3(x-8)&=&2\\\log_3({\color{blue}{9}})+\log_3({\color{blue}{9}}-8)&\overset?=&2\\2+0&\overset?=&2\\2&=&2\;\checkmark\end{eqnarray*}[/latex]

Step 1: Use the Quotient Property on the left side and the Power Property on the right.

[latex]\log_4\left(\frac{x+6}{2x+5}\right)=\log_4x^{-1}[/latex]

Step 2: Rewrite [latex]{x}^{-1}=\frac{1}{x}[/latex].

[latex]\log_4\left(\frac{x+6}{2x+5}\right)=\log_4x^\frac14[/latex]

Step 3: Use the One-to-One Property, if [latex]\log_aM=\log_aN[/latex], then [latex]M=N[/latex].

[latex]\frac{x+6}{2x+5}=\frac1x[/latex]

Step 4: Solve the rational equation.

[latex]x(x+6)=2x+5[/latex]

Step 5: Distribute.

[latex]x^2+6x=2x+5[/latex]

Step 6: Write in standard form. 

[latex]x^2+4x-5=0[/latex]

Step 7: Factor.

[latex](x+5)(x-1)=0[/latex]

Step 8: Use the Zero-Product Property.

[latex]x+5=0,\;x-1=0[/latex]

Step 9: Solve each equation.

[latex]\cancel{x=-5},\;\;x=1[/latex]

Step 10: Check. 

We leave the check for you.

Step 1: Since the exponential is isolated, take the logarithm of both sides.

[latex]\log5^x=\log11[/latex]

Step 2: Use the Power Property to get the [latex]x[/latex] as a factor, not an exponent. 

[latex]x\log5=\log11[/latex]

Step 3: Solve for [latex]x[/latex]. Find the exact answer. 

[latex]x=\frac{\log11}{\log5}[/latex]

Step 4: Approximate the answer. 

[latex]x\approx1.490[/latex]

Step 5: Check — Does it make sense?

Since [latex]5^1=5[/latex], [latex]5^2=25[/latex] does it make sense that [latex]5^{1.490}\approx11[/latex]?

Step 1: Isolate the exponential by dividing both sides by [latex]3[/latex].

[latex]e^{x+2}=8[/latex]

Step 2: Take the natural logarithm of both sides. 

[latex]\textstyle\text{ln}\hspace{0.2em}e^{x+2}\hspace{0.2em}=\hspace{0.2em}\text{ln}8\hspace{1.8em}[/latex]

Step 3: Use the Power Property to get the [latex]x[/latex] as a factor, not an exponent.

[latex]\textstyle{(x+2)}\text{ln}\hspace{0.2em}e\hspace{0.2em}=\hspace{0.2em}\text{ln}8\hspace{1.8em}[/latex]

Step 4: Use the property [latex]\ln\;e=1[/latex] to simplify.

[latex]\textstyle x+2\hspace{0.2em}=\hspace{0.2em}\text{ln}8\hspace{1.8em}[/latex]

Step 5: Solve the equation. Find the exact answer. 

[latex]\textstyle x\hspace{0.2em}=\hspace{0.2em}\text{ln}8-2[/latex]

Step 6: Approximate the answer. 

[latex]\textstyle x\approx0.079[/latex]

Step 1: Identify the variables in the formula.

[latex]\begin{eqnarray*}\textstyle A&=&\$50,000\\P&=&\$10,000\\r&=&?\\t&=&17\;\text{years}\\A&=&Pe^{rt}\end{eqnarray*}[/latex]

Step 2: Substitute the values into the formula. 

[latex]\textstyle50,000=10,000e^{r\cdot17}[/latex]

Step 3: Solve for [latex]r[/latex]. Divide each side by 10,000.

[latex]\textstyle5=e^{17r}[/latex]

Step 4: Take the natural log of each side. 

[latex]\textstyle\text{ln}5=\text{ln}\hspace{0.2em}e^{17r}[/latex]

Step 5: Use the Power Property. 

[latex]\textstyle\text{ln}5=17r\text{ln}\hspace{0.2em}e[/latex]

Step 6: Simplify. 

[latex]\textstyle\text{ln}5=17r[/latex]

Step 7: Divide each side by [latex]17[/latex]. 

[latex]\textstyle\frac{\text{ln}5}{17}=r[/latex]

Step 8: Approximate the answer. 

[latex]\textstyle r\approx0.095[/latex]

Step 9: Convert to a percentage. 

[latex]\textstyle r\approx9.5\%[/latex]

They need the rate of growth to be approximately [latex]9.5\%[/latex].

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

Step 1: Identify the variables in the formula. 

[latex]\begin{eqnarray*}\textstyle A&=&300\\A_0&=&100\\k&=&?\\t&=&3\;\text{hours}\\A&=&A_0e^{kt}\end{eqnarray*}[/latex]

Step 2: Substitute the values in the formula. 

[latex]\textstyle\hspace{0.1em}300=100e^{k\cdot3}[/latex]

Step 3: Solve for [latex]k[/latex]. Divide each side by [latex]100[/latex]. 

[latex]\textstyle3=e^{3k}[/latex]

Step 4: Take the natural log of each side. 

[latex]\textstyle\text{ln}3=\text{ln}\hspace{0.2em}e^{3k}[/latex]

Step 5: Use the Power Property. 

[latex]\textstyle\text{ln}3=3k\text{ln}\hspace{0.2em}e[/latex]

Step 6: Simplify. 

[latex]\textstyle\text{ln}3=3k[/latex]

Step 7: Divide each side by [latex]3[/latex].

[latex]\textstyle\frac{\text{ln}3}3=k[/latex]

Step 8: Approximate the answer. 

[latex]\textstyle k\approx0.366[/latex]

We use this rate of growth to predict the number of bacteria there will be in [latex]24[/latex] hours.

[latex]\begin{eqnarray*}\textstyle A&=&?\\A_0&=&100\\k&=&{\displaystyle\frac{\ln3}3}\\t&=&24\;\text{hours}\\A&=&A_0e^{kt}\end{eqnarray*}[/latex]

Step 9: Substitute in the values. 

[latex]\textstyle A=100e^{\frac{\text{ln}3}3\cdot24}[/latex]

Step 10: Evaluate. 

[latex]\textstyle A\approx656,100[/latex]

At this rate of growth, they can expect [latex]656,10[/latex]0 bacteria.

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

Step 1: Identify the variables in the formula. 

[latex]\begin{eqnarray*}\textstyle A&=&50\\A_0&=&100\\k&=&?\\t&=&\;1590\;\text{years}\\A&=&A_0e^{kt}\end{eqnarray*}[/latex]

Step 2: Substitute the values in the formula. 

[latex]\textstyle50=100e^{k\cdot1590}[/latex]

Step 3: Solve for [latex]k[/latex]. Divide each side by [latex]100[/latex]. 

[latex]\textstyle0.5=e^{1590k}[/latex]

Step 4: Take the natural log of each side. 

[latex]\textstyle\text{ln}0.5=\text{ln}\hspace{0.2em}e^{1590k}[/latex]

Step 5: Use the Power Property. 

[latex]\textstyle\text{ln}0.5=1590k\text{ln}\hspace{0.2em}e[/latex]

Step 6: Simplify.

[latex]\textstyle\text{ln}0.5=1590k[/latex]

Step 7: Divide each side by [latex]1590[/latex].

[latex]\textstyle\frac{\text{ln}0.5}{1590}=k\hspace{0.2em}\text{exact answer}[/latex]

Step 8: Substitute in the values. 

[latex]\textstyle A=100e^{\frac{\text{ln}0.5}{1590}\cdot500}[/latex]

We use this rate of growth to predict the amount that will be left in [latex]500[/latex] years.

[latex]\begin{eqnarray*}\textstyle A&=&?\\A_0&=&100\\k&=&{\displaystyle\frac{\ln0.5}{1590}}\\t&=&500\;\text{years}\\A&=&A_0e^{kt}\end{eqnarray*}[/latex]

Step 9: Evaluate. 

[latex]\textstyle A\approx80.4\hspace{0.2em}\text{mg}[/latex]

In [latex]500[/latex] years there would be approximately [latex]80.4[/latex] mg remaining.