In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.
Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section we will learn techniques for solving exponential and logarithmic equations.
Learning Outcomes
- Solve an exponential equation using the definition of a logarithm.
- Solve an exponential equation graphically.
- Solve a logarithmic equation graphically.
Exponential Equations
The first technique we will introduce for solving exponential equations involves using the definition of the logarithm.
How To: Given an equation of the form [latex]y=A{e}^{kt}[/latex], solve for [latex]t[/latex]
- Isolate the exponential expression, that is, the base with its exponent should be isolated to one side of the equation.
- Change from exponential form of the equation to logarithmic form.
- Use your calculator to find the approximate solution.
Example: Solving an Equation of the Form [latex]y=A{e}^{kt}[/latex]
Solve [latex]100=20{e}^{2t}[/latex].
Try It
Solve [latex]3{e}^{0.5t}=11[/latex].
Example: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[/latex]
Solve [latex]4{e}^{2x}+5=12[/latex].
Try It
Solve [latex]3+{e}^{2t}=7{e}^{2t}[/latex].
Logarithmic Equations
We have already seen that every logarithmic equation [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex] is equal to the exponential equation [latex]{b}^{y}=x[/latex]. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.
A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations
For any algebraic expression S and real numbers b and c, where [latex]b>0,\text{ }b\ne 1[/latex],
[latex]{\mathrm{log}}_{b}\left(S\right)=c\text{ if and only if }{b}^{c}=S[/latex]
Example: Using Algebra to Solve a Logarithmic Equation
Solve [latex]2\mathrm{ln}x+3=7[/latex].
Try It
Solve [latex]6+\mathrm{ln}x=10[/latex].
Example: Using Algebra Before and After Using the Definition of the Natural Logarithm
Solve [latex]2\mathrm{ln}\left(6x\right)=7[/latex].
Try It
Solve [latex]2\mathrm{ln}\left(x+1\right)=10[/latex].
Example: Using a Graph to Understand the Solution to a Logarithmic Equation
Solve [latex]\mathrm{ln}x=3[/latex].
Try It
Use a graphing calculator to estimate the approximate solution to the logarithmic equation [latex]{2}^{x}=1000[/latex] to 2 decimal places.
Change-of-Base Formula for Logarithms
Some calculators can only evaluate common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[/latex], we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.
To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms.
Given any positive real numbers M, b, and n, where [latex]n\ne 1[/latex] and [latex]b\ne 1[/latex], we show
[latex]{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}[/latex]
Let [latex]y={\mathrm{log}}_{b}M[/latex]. Converting to exponential form, we obtain [latex]{b}^{y}=M[/latex]. It follows that:
[latex]\begin{array}{l}{\mathrm{log}}_{n}\left({b}^{y}\right)\hfill & ={\mathrm{log}}_{n}M\hfill & \text{Apply the one-to-one property}.\hfill \\ y{\mathrm{log}}_{n}b\hfill & ={\mathrm{log}}_{n}M \hfill & \text{Apply the power rule for logarithms}.\hfill \\ y\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Isolate }y.\hfill \\ {\mathrm{log}}_{b}M\hfill & =\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\hfill & \text{Substitute for }y.\hfill \end{array}[/latex]
For example, to evaluate [latex]{\mathrm{log}}_{5}36[/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.
[latex]\begin{array}{l}{\mathrm{log}}_{5}36\hfill & =\frac{\mathrm{log}\left(36\right)}{\mathrm{log}\left(5\right)}\hfill & \text{Apply the change of base formula using base 10}\text{.}\hfill \\ \hfill & \approx 2.2266\text{ }\hfill & \text{Use a calculator to evaluate to 4 decimal places}\text{.}\hfill \end{array}[/latex]
A General Note: The Change-of-Base Formula
The change-of-base formula can be used to evaluate a logarithm with any base.
For any positive real numbers M, b, and n, where [latex]n\ne 1[/latex] and [latex]b\ne 1[/latex],
[latex]{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}[/latex].
It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.
[latex]{\mathrm{log}}_{b}M=\frac{\mathrm{ln}M}{\mathrm{ln}b}[/latex]
and
[latex]{\mathrm{log}}_{b}M=\frac{\mathrm{log}M}{\mathrm{log}b}[/latex]
Q & A
Can we change common logarithms to natural logarithms?
Yes. Remember that [latex]\mathrm{log}9[/latex] means [latex]{\text{log}}_{\text{10}}\text{9}[/latex]. So, [latex]\mathrm{log}9=\frac{\mathrm{ln}9}{\mathrm{ln}10}[/latex].
Example: Using the Change-of-Base Formula with a Calculator
Evaluate [latex]{\mathrm{log}}_{2}\left(10\right)[/latex] using the change-of-base formula with a calculator.
Try It
Evaluate [latex]{\mathrm{log}}_{5}\left(100\right)[/latex] using the change-of-base formula.
- We can convert a logarithm with any base to a quotient of logarithms with any other base using the change-of-base formula.
- The change-of-base formula is often used to rewrite a logarithm with a base other than 10 or [latex]e[/latex] as the quotient of natural or common logs. A calculator can then be used to evaluate it.
Using a Graph to Approximate a Solution to an Exponential Equation
Graphing can help you confirm or find the solution to an exponential equation. For example,[latex]42=1.2{\left(5\right)}^{x}+2.8[/latex] can be solved to find the specific value for x that makes it a true statement. Graphing [latex]y=4[/latex] along with [latex]y=2^{x}[/latex] in the same window, the point(s) of intersection if any represent the solutions of the equation.
To use a calculator to solve this, press [Y=] and enter [latex]1.2(5)x+2.8[/latex] next to Y1=. Then enter 42 next to Y2=. For a window, use the values –3 to 3 for[latex]x[/latex] and –5 to 55 for[latex]y[/latex].Press [GRAPH]. The graphs should intersect somewhere near[latex]x=2[/latex].
For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth,x≈2.166.
Try It
Solve [latex]4=7.85{\left(1.15\right)}^{x}-2.27[/latex] graphically. Round to the nearest thousandth.
Using a Graph to Approximate a Solution to a Logarithmic Equation
How To: Given a logarithmic equation, use a graphing calculator to approximate solutions
- Press [Y=]. Enter the given logarithmic equation or equations as Y1= and, if needed, Y2=.
- Press [GRAPH] to observe the graphs of the curves and use [WINDOW] to find an appropriate view of the graphs, including their point(s) of intersection.
- To find the value of x, we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x for the point(s) of intersection.
Example: Approximating the Solution of a Logarithmic Equation
Solve [latex]4\mathrm{ln}\left(x\right)+1=-2\mathrm{ln}\left(x - 1\right)[/latex] graphically. Round to the nearest thousandth.
Try It
Solve [latex]5\mathrm{log}\left(x+2\right)=4-\mathrm{log}\left(x\right)[/latex] graphically. Round to the nearest thousandth.
Key Concepts
- When given an equation of the form [latex]{\mathrm{log}}_{b}\left(S\right)=c[/latex], where S is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation [latex]{b}^{c}=S[/latex] and solve for the unknown.
- A graphing calculator may be used to approximate solutions to some exponential and logarithmic equations.
- We can also use graphing to solve equations of the form [latex]{\mathrm{log}}_{b}\left(S\right)=c[/latex]. We graph both equations [latex]y={\mathrm{log}}_{b}\left(S\right)[/latex] and [latex]y=c[/latex] on the same coordinate plane and identify the solution as the x-value of the point of intersecting.
- Combining the skills learned in this and previous sections, we can solve equations that model real world situations whether the unknown is in an exponent or in the argument of a logarithm.
Glossary
change-of-base formula
a formula for converting a logarithm with any base to a quotient of logarithms with any other base
Definition of a logarithm | For any algebraic expression S and positive real numbers b and c, where [latex]b\ne 1[/latex], [latex]{\mathrm{log}}_{b}\left(S\right)=c[/latex] if and only if [latex]{b}^{c}=S[/latex]. |
The Change-of-Base Formula | [latex]{\mathrm{log}}_{b}M\text{=}\frac{{\mathrm{log}}_{n}M}{{\mathrm{log}}_{n}b}\text{ }n>0,n\ne 1,b\ne 1[/latex] |
Candela Citations
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