What Is the Approximate Solution of E2x 3 7
Learning Outcomes
- Solve an exponential equation with a common base.
- Rewrite an exponential equation so all terms have a common base then solve.
- Recognize when an exponential equation does not have a solution.
- Use logarithms to solve exponential equations.
Exponential Equations
The first technique we will introduce for solving exponential equations involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where [latex]b>0,\text{ }b\ne 1[/latex], [latex]{b}^{S}={b}^{T}[/latex] if and only if S= T.
In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
For example, consider the equation [latex]{3}^{4x - 7}=\frac{{3}^{2x}}{3}[/latex]. To solve for x, we use the division property of exponents to rewrite the right side so that both sides have the common base 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x:
[latex]\begin{array}{l}{3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{3}\hfill & \hfill \\ {3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{{3}^{1}}\hfill & {\text{Rewrite 3 as 3}}^{1}.\hfill \\ {3}^{4x - 7}\hfill & ={3}^{2x - 1}\hfill & \text{Use the division property of exponents}\text{.}\hfill \\ 4x - 7\hfill & =2x - 1\text{ }\hfill & \text{Apply the one-to-one property of exponents}\text{.}\hfill \\ 2x\hfill & =6\hfill & \text{Subtract 2}x\text{ and add 7 to both sides}\text{.}\hfill \\ x\hfill & =3\hfill & \text{Divide by 3}\text{.}\hfill \end{array}[/latex]
A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations
For any algebraic expressions S and T, and any positive real number [latex]b\ne 1[/latex],
[latex]{b}^{S}={b}^{T}\text{ if and only if }S=T[/latex]
How To: Given an exponential equation Of the form [latex]{b}^{S}={b}^{T}[/latex], where S andT are algebraic expressions with an unknown, solve for the unknown
- Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[/latex].
- Use the one-to-one property to set the exponents equal to each other.
- Solve the resulting equation, S= T, for the unknown.
Example: Solving an Exponential Equation with a Common Base
Solve [latex]{2}^{x - 1}={2}^{2x - 4}[/latex].
Show Solution
[latex]\begin{array}{l} {2}^{x - 1}={2}^{2x - 4}\hfill & \text{The common base is }2.\hfill \\ \text{ }x - 1=2x - 4\hfill & \text{By the one-to-one property the exponents must be equal}.\hfill \\ \text{ }x=3\hfill & \text{Solve for }x.\hfill \end{array}[/latex]
Try It
Solve [latex]{5}^{2x}={5}^{3x+2}[/latex].
Show Solution
[latex]x=–2[/latex]
Rewriting Equations So All Powers Have the Same Base
Sometimes the common base for an exponential equation is not explicitly shown. In these cases we simply rewrite the terms in the equation as powers with a common base and solve using the one-to-one property.
For example, consider the equation [latex]256={4}^{x - 5}[/latex]. We can rewrite both sides of this equation as a power of 2. Then we apply the rules of exponents along with the one-to-one property to solve for x:
[latex]\begin{array}{l}256={4}^{x - 5}\hfill & \hfill \\ {2}^{8}={\left({2}^{2}\right)}^{x - 5}\hfill & \text{Rewrite each side as a power with base 2}.\hfill \\ {2}^{8}={2}^{2x - 10}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ 8=2x - 10\hfill & \text{Apply the one-to-one property of exponents}.\hfill \\ 18=2x\hfill & \text{Add 10 to both sides}.\hfill \\ x=9\hfill & \text{Divide by 2}.\hfill \end{array}[/latex]
How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it
- Rewrite each side in the equation as a power with a common base.
- Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[/latex].
- Use the one-to-one property to set the exponents equal to each other.
- Solve the resulting equation, S= T, for the unknown.
Example: Solving Equations by Rewriting Them to Have a Common Base
Solve [latex]{8}^{x+2}={16}^{x+1}[/latex].
Show Solution
[latex]\begin{array}{llllll}\text{ }{8}^{x+2}={16}^{x+1}\hfill & \hfill \\ {\left({2}^{3}\right)}^{x+2}={\left({2}^{4}\right)}^{x+1}\hfill & \text{Write }8\text{ and }16\text{ as powers of }2.\hfill \\ \text{ }{2}^{3x+6}={2}^{4x+4}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ \text{ }3x+6=4x+4\hfill & \text{Use the one-to-one property to set the exponents equal to each other}.\hfill \\ \text{ }x=2\hfill & \text{Solve for }x.\hfill \end{array}[/latex]
Try It
Solve [latex]{5}^{2x}={25}^{3x+2}[/latex].
Show Solution
[latex]x=–1[/latex]
Example: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base
Solve [latex]{2}^{5x}=\sqrt{2}[/latex].
Show Solution
[latex]\begin{array}{l}{2}^{5x}={2}^{\frac{1}{2}}\hfill & \text{Write the square root of 2 as a power of }2.\hfill \\ 5x=\frac{1}{2}\hfill & \text{Use the one-to-one property}.\hfill \\ x=\frac{1}{10}\hfill & \text{Solve for }x.\hfill \end{array}[/latex]
Try It
Solve [latex]{5}^{x}=\sqrt{5}[/latex].
Show Solution
[latex]x=\frac{1}{2}[/latex]
Q & A
Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?
No. Recall that the range of an exponential function is always positive. While solving the equation we may obtain an expression that is undefined.
Example: Determine When an Equation has No Solution
Solve [latex]{3}^{x+1}=-2[/latex].
Try It
Solve [latex]{2}^{x}=-100[/latex].
Show Solution
The equation has no solution.
Using Logarithms to Solve Exponential Equations
Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall that since [latex]\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)[/latex] is equal to a= b,we may apply logarithms with the same base to both sides of an exponential equation.
How To: Given an exponential equation Where a common base cannot be found, solve for the unknown
- Apply the logarithm to both sides of the equation.
- If one of the terms in the equation has base 10, use the common logarithm.
- If none of the terms in the equation has base 10, use the natural logarithm.
- Use the rules of logarithms to solve for the unknown.
Example: Solving an Equation Containing Powers of Different Bases
Solve [latex]{5}^{x+2}={4}^{x}[/latex].
Show Solution
[latex]\begin{array}{l}\text{ }{5}^{x+2}={4}^{x}\hfill & \text{There is no easy way to get the powers to have the same base}.\hfill \\ \text{ }\mathrm{ln}{5}^{x+2}=\mathrm{ln}{4}^{x}\hfill & \text{Take ln of both sides}.\hfill \\ \text{ }\left(x+2\right)\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the power rule for logs}.\hfill \\ \text{ }x\mathrm{ln}5+2\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the distributive property}.\hfill \\ \text{ }x\mathrm{ln}5-x\mathrm{ln}4=-2\mathrm{ln}5\hfill & \text{Get terms containing }x\text{ on one side, terms without }x\text{ on the other}.\hfill \\ x\left(\mathrm{ln}5-\mathrm{ln}4\right)=-2\mathrm{ln}5\hfill & \text{On the left hand side, factor out }x.\hfill \\ \text{ }x\mathrm{ln}\left(\frac{5}{4}\right)=\mathrm{ln}\left(\frac{1}{25}\right)\hfill & \text{Use the properties of logs}.\hfill \\ \text{ }x=\frac{\mathrm{ln}\left(\frac{1}{25}\right)}{\mathrm{ln}\left(\frac{5}{4}\right)}\hfill & \text{Divide by the coefficient of }x.\hfill \end{array}[/latex]
Try It
Solve [latex]{2}^{x}={3}^{x+1}[/latex].
Show Solution
[latex]x=\frac{\mathrm{ln}3}{\mathrm{ln}\left(\frac{2}{3}\right)}[/latex]
Q & A
Is there any way to solve [latex]{2}^{x}={3}^{x}[/latex]?
Yes. The solution is x = 0.
Equations Containing [latex]e[/latex]
One common type of exponential equations are those with base e. This constant occurs again and again in nature, mathematics, science, engineering, and finance. When we have an equation with a base e on either side, we can use the natural logarithm to solve it.
How To: Given an equation of the form [latex]y=A{e}^{kt}[/latex], solve for [latex]t[/latex]
- Divide both sides of the equation by A.
- Apply the natural logarithm to both sides of the equation.
- Divide both sides of the equation by k.
TIPS for Success
Just as you have done when solving various types of equations, isolate the term containing the variable for which you are solving before applying any properties of equality or inverse operations. That's why, in the example above, you must divide away the [latex]A[/latex] first.
Remember that the functions [latex]y=e^{x}[/latex] and [latex]y=\mathrm{ln}\left(x\right)[/latex] are inverse functions.
Therefore, [latex]\mathrm{ln}\left({e}^{x}\right)=x[/latex] for all [latex]x[/latex] and [latex]e^{\mathrm{ln}\left(x\right)}=x[/latex] for [latex]x>0[/latex].
Example: Solve 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].
Show Solution
[latex]t=2\mathrm{ln}\left(\frac{11}{3}\right)[/latex] or [latex]\mathrm{ln}{\left(\frac{11}{3}\right)}^{2}[/latex]
Q & A
Does every equation of the form [latex]y=A{e}^{kt}[/latex] have a solution?
No. There is a solution when [latex]k\ne 0[/latex], and when [latex]y[/latex] and [latex]A[/latex] are either both 0 or neither 0 and they have the same sign. An example of an equation with this form that has no solution is [latex]2=-3{e}^{t}[/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].
Show Solution
[latex]\begin{array}{l}4{e}^{2x}+5=12\hfill & \hfill \\ 4{e}^{2x}=7\hfill & \text{Subtract 5 from both sides}.\hfill \\ {e}^{2x}=\frac{7}{4}\hfill & \text{Divide both sides by 4}.\hfill \\ 2x=\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Take ln of both sides}.\hfill \\ x=\frac{1}{2}\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Solve for }x.\hfill \end{array}[/latex]
Try It
Solve [latex]3+{e}^{2t}=7{e}^{2t}[/latex].
Show Solution
[latex]t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)[/latex]
Extraneous Solutions
Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when taking the logarithm of both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.
Example: Solving Exponential Functions in Quadratic Form
Solve [latex]{e}^{2x}-{e}^{x}=56[/latex].
Try It
Solve [latex]{e}^{2x}={e}^{x}+2[/latex].
Show Solution
[latex]x=\mathrm{ln}2[/latex]
Q & A
Does every logarithmic equation have a solution?
No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.
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What Is the Approximate Solution of E2x 3 7
Source: https://courses.lumenlearning.com/waymakercollegealgebra/chapter/exponential-equations-with-like-bases/