4.6: Exponential and Logarithmic Equations
In this section, you will:
- Use like bases to solve exponential equations.
- Use logarithms to solve exponential equations.
- Use the definition of a logarithm to solve logarithmic equations.
- Use the One-to-One Property of logarithms to solve logarithmic equations.
- Solve applied problems involving exponential and logarithmic equations.
Figure \( \PageIndex{ 1 } \): Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the "rabbit plague." (credit: Richard Taylor, Flickr)
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 functions.
Using Like Bases to Solve Exponential Equations
The first technique involves two functions with like bases. Recall the following theorem from Algebra.
For any algebraic expressions \(S\) and \(T\), and any positive real number \(b \neq 1\),\[b^S = b^T \text{ if and only if } S = T. \nonumber \]
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 laws 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 \( 3^{4x - 7} = \frac{3^{2x}}{3} \). To solve for \(x\), we use the quotient 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 \(x\):\[ \begin{array}{rrclcl} & 3^{4x - 7} & = & \dfrac{3^{2x}}{3} & & \\[6pt] \implies & 3^{4x - 7} & = & \dfrac{3^{2x}}{3^1} & \quad & \left( 3 = 3^1 \right) \\[6pt] \implies & 3^{4x - 7} & = & 3^{2x - 1} & \quad & \left( \text{Laws of Exponents: Quotient Rule} \right) \\[6pt] \implies & 4x - 7 & = & 2x - 1 & \quad & \left( \text{One-to-One Property of Exponents} \right) \\[6pt] \implies & 2x & = & 6 & \quad & \left( \text{subtracting }2x\text{ and adding }7\text{ to both sides} \right) \\[6pt] \implies & x & = & 3 & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \end{array} \nonumber \]
Solve \(2^{x−1} = 2^{2x−4} \).
- Solution
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\[ \begin{array}{rrclcl} & 2^{x - 1} & = & 2^{2x - 4} & & \\[6pt] \implies & x - 1 & = & 2x - 4 & \quad & \left( \text{One-to-One Property} \right) \\[6pt] \implies & 3 & = & x & \quad & \left( \text{subtracting }x\text{ and adding }4\text{ to both sides} \right) \\[6pt] \end{array} \nonumber \]
Solve \(5^{2x} = 5^{3x+2} \).
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 \(256 = 4^{x−5} \). We can rewrite both sides of this equation as a power of \(2\). Then we apply the Laws of Exponents, along with the One-to-One Property, to solve for \(x\):\[ \begin{array}{rrclcl} & 256 & = & 4^{x - 5} & & \\[6pt] \implies & 4^4 & = & 4^{x - 5} & \quad & \left( \text{rewriting both sides in the same base} \right) \\[6pt] \implies & 4 & = & x - 5 & \quad & \left( \text{One-to-One Property} \right) \\[6pt] \implies & 9 & = & x & \quad & \left( \text{adding }5\text{ to both sides} \right) \\[6pt] \end{array} \nonumber \]
Solve \(8^{x+2} = 16^{x+1} \).
- Solution
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\[ \begin{array}{rrclcl} & 8^{x + 2} & = & 16^{x + 1} & & \\[6pt] \implies & \left( 2^3 \right)^{x + 2} & = & \left( 2^4 \right)^{x + 1} & \quad & \left( \text{rewriting both sides with the same base} \right) \\[6pt] \implies & 2^{3(x + 2)} & = & 2^{4(x + 1)} & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] \implies & 2^{3x + 6} & = & 2^{4x + 4} & \quad & \left( \text{distributing} \right) \\[6pt] \implies & 3x + 6 & = & 4x + 4 & \quad & \left( \text{One-to-One Property} \right) \\[6pt] \implies & 2 & = & x & \quad & \left( \text{subtracting }3x\text{ and }4\text{ from both sides} \right) \\[6pt] \end{array} \nonumber \]
Solve \(5^{2x} = 25^{3x+2} \).
Solve \(2^{5x} =\sqrt{ 2 } \).
- Solution
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\[ \begin{array}{rrclcl} & 2^{5x} & = & \sqrt{2} & & \\[6pt] \implies & 2^{5x} & = & 2^{1/2} & \quad & \left( \text{rewriting the radical as a rational exponent} \right) \\[6pt] \implies & 5x & = & \dfrac{1}{2} & \quad & \left( \text{One-to-One Property} \right) \\[6pt] \implies & x & = & \dfrac{1}{10} & \quad & \left( \text{dividing both sides by }5 \right) \\[6pt] \end{array} \nonumber \]
Solve \(5^x =\sqrt{ 5 } \).
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.
Solve \(3^{x+1} =−2\).
- Solution
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This equation has no solution. There is no real value of \(x\) that will make the equation a true statement because any power of a positive number is positive.
Figure \( \PageIndex{ 2 } \) shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.
Figure \( \PageIndex{ 2 } \)
Solve \(2^x =−100\).
Solving Exponential Equations Using Logarithms
Most of the time 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, since \( \log( a ) = \log( b )\) is equivalent to \(a=b\), we may apply logarithms with the same base on both sides of an exponential equation.
Solve \(5^{x + 2} = 4^x \).
- Solution
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\[ \begin{array}{rrclcl} & 5^{x + 2} & = & 4^x & \quad & \left( \text{notice there is no common base between }5\text{ and }4 \right) \\[6pt] \implies & \ln\left( 5^{x + 2} \right) & = & \ln\left( 4^x \right) & \quad & \left( \text{take the natural log of both sides} \right) \\[6pt] \implies & (x + 2)\ln\left( 5 \right) & = & x \ln\left( 4 \right) & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] \implies & x\ln(5) + 2\ln(5) & = & x \ln(4) & \quad & \left( \text{distributing} \right) \\[6pt] \implies & x\ln(5) - x\ln(4) & = & -2 \ln(5) & \quad & \left( \text{moving all terms containing}x\text{ to one side} \right) \\[6pt] \implies & x(\ln(5) - \ln(4)) & = & -2 \ln(5) & \quad & \left( \text{factoring out }x \right) \\[6pt] \implies & x & = & \dfrac{-2 \ln(5)}{\ln(5) - \ln(4)} & \quad & \left( \text{dividing both sides by }\ln(5) - \ln(4) \right) \\[6pt] \end{array} \nonumber \]Some could correctly argue that we could further simplify the final expression using the Laws of Logarithms.\[ x = \dfrac{-2 \ln(5)}{\ln(5) - \ln(4)} = \dfrac{\ln\left( 5^{-2} \right)}{\ln\left( \frac{5}{4} \right)}. \nonumber \]However, this is as correct as \( \frac{-2 \ln(5)}{\ln(5) - \ln(4)} \). In fact, you could go even further in your simplification to\[ x = \dfrac{\ln\left( 5^{-2} \right)}{\ln\left( \frac{5}{4} \right)} = \dfrac{\ln\left( \frac{1}{25} \right)}{\ln\left( \frac{5}{4} \right)}. \nonumber \]Be sure to check with your instructor as to their preference, but my guess is that any of these is acceptable as equivalent.
Solve \(2^x = 3^{x+1} \).
Is there any way to solve \(2^x = 3^x \) without logarithms?
Yes. The solution is \(0\).
Equations Containing \( e \)
One common type of exponential equations are those with base \(e\). This constant occurs again and again in nature, in Mathematics, in Science, in Engineering, and in finance. When we have an equation with a base \(e\) on either side, we can use the natural logarithm to solve it.
Solve \(100 = 20 e^{2t} \).
- Solution
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\[ \begin{array}{rrclcl} & 100 & = & 20 e^{2t} & & \\[6pt] \implies & 5 & = & e^{2t} & \quad & \left( \text{dividing both sides by }20 \right) \\[6pt] \implies & \ln(5) & = & \ln\left(e^{2t}\right) & \quad & \left( \text{taking the natural log of both sides} \right) \\[6pt] \implies & \ln(5) & = & 2t & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] \implies & \dfrac{\ln(5)}{2} & = & t & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \end{array} \nonumber \]Using Laws of Logarithms, we can also write this answer in the form \(t = \ln\sqrt{ 5 } \). If we want a decimal approximation of the answer, we use a calculator.
Solve \(3 e^{0.5t} =11\).
Does every equation of the form \(y=A e^{kt}\) have a solution?
No. There is a solution when \(k \neq 0\), and when \(y\) and \(A\) 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 \(2 = −3 e^t \).
Solve \(4 e^{2x} +5=12\).
- Solution
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\[ \begin{array}{rrclcl} & 4 e^{2x} + 5 & = & 12 & & \\[6pt] \implies & 4 e^{2x} & = & 7 & \quad & \left( \text{subtracting }5\text{ from both sides} \right) \\[6pt] \implies & e^{2x} & = & \dfrac{7}{4} & \quad & \left( \text{dividing both sides by }4 \right) \\[6pt] \implies & \ln\left(e^{2x}\right) & = & \ln\left(\dfrac{7}{4}\right) & \quad & \left( \text{taking the natural log of both sides} \right) \\[6pt] \implies & 2x & = & \ln\left(\dfrac{7}{4}\right) & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] \implies & x & = & \dfrac{\ln\left(7/4\right)}{2} & \quad & \left( \text{dividing both sides by}2 \right) \\[6pt] \end{array} \nonumber \]
Solve \(3+ e^{2t} =7 e^{2t} \).
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 the logarithm is taken on 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.
Solve \(e^{2x} − e^x = 56\).
- Solution
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\[ \begin{array}{rrclcl} & e^{2x} - e^x & = & 56 & & \\[6pt] \implies & e^{2x} - e^x - 56 & = & 0 & \quad & \left( \text{subtracting }56\text{ from both sides} \right) \\[6pt] \implies & \left(e^{x}\right)^2 - e^x - 56 & = & 0 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] \implies & u^2 - u - 56 & = & 0 & \quad & \left( \text{substituting }u = e^x \right) \\[6pt] \implies & (u - 8)(u + 7) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt] \end{array} \nonumber \]Therefore, \( u = 8 \) or \( u = -7 \). Resubstituting \( u = e^x \), we get\[ e^x = 8 \quad \text{or} \quad e^x = -7. \nonumber \]Since \( e^x > 0 \), we throw out the possibility that \( e^x = -7 \). To solve \( e^x = 8 \), we take the natural log of both sides to get \( x = \ln(8) \).
When we plan to use factoring to solve a problem, as we did in Example \( \PageIndex{ 8 } \), we always get zero on one side of the equation. This is because zero has the unique property that when a product is zero, one or both of the factors must be zero. We rejected the equation \(e^x =−7\) because a positive number never equals a negative number. The solution \( \ln(−7)\) is not a real number, and in the real number system this solution is rejected as an extraneous solution.
Solve \(e^{2x} = e^x +2\).
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.
Using the Definition of a Logarithm to Solve Logarithmic Equations
We have already seen that every logarithmic equation \( \log_b(x) = y \) is equivalent to the exponential equation \( b^y = x \). We can use this fact, along with the Laws of Logarithms, to solve logarithmic equations where the argument is an algebraic expression.
For example, consider the equation \( \log_2(2) + \log_2(3x - 5) = 3 \). To solve this equation, we can use Laws of Logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for \(x\):\[ \begin{array}{rrclcl} & \log_2(2) + \log_2(3x - 5) & = & 3 & & \\[6pt] \implies & \log_2(2(3x - 5)) & = & 3 & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] \implies & \log_2(6x - 10) & = & 3 & \quad & \left( \text{distributing} \right) \\[6pt] \implies & 6x - 10 & = & 2^3 & \quad & \left( \text{definition of a logarithm} \right) \\[6pt] \implies & 6x - 10 & = & 8 & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & 6x & = & 18 & \quad & \left( \text{adding }10\text{ to both sides} \right) \\[6pt] \implies & x & = & 3 & \quad & \left( \text{dividing both sides by }6 \right) \\[6pt] \end{array} \nonumber \]As you can see, when solving a logarithmic equation, we rely on the following definition of logarithms:\[ \text{For any algebraic expression } S \text{ and real numbers } b \text{ and } c, \text{ where } b > 0 \text{ and } b \neq 1, \log_b (S) = c \text{ if and only if } b^c = S. \nonumber \]
Solve \(2 \ln (x) + 3 = 7\).
- Solution
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\[ \begin{array}{rrclcl} & 2\ln(x) + 3 & = & 7 & & \\[6pt] \implies & 2 \ln(x) & = & 4 & \quad & \left( \text{subtracting }3\text{ from both sides} \right) \\[6pt] \implies & \ln(x) & = & 2 & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \implies & x & = & e^2 & \quad & \left( \text{rewriting in exponential form} \right) \\[6pt] \end{array} \nonumber \]
Solve \(6 + \ln x = 10\).
Solve \(2 \ln(6x)=7\).
- Solution
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\[ \begin{array}{rrclcl} & 2 \ln(6x) & = & 7 & & \\[6pt] \implies & \ln(6x) & = & \dfrac{7}{2} & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt] \implies & 6x & = & e^{7/2} & \quad & \left( \text{rewriting in exponential form} \right) \\[6pt] \implies & x & = & \dfrac{1}{6} e^{7/2} & \quad & \left( \text{dividing both sides by }6 \right) \\[6pt] \end{array} \nonumber \]
Solve \(2 \ln(x+1)=10\).
Solve \( \log_7(1- 2x) = 1 - \log_7(3 - x) \).
- Solution
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\[ \begin{array}{rrclcl} & \log_7(1 - 2x) & = & 1 - \log_7(3 - x) & & \\[6pt] \implies & \log_7(1 - 2x) + \log_7(3 - x) & = & 1 & \quad & \left( \text{adding }\log_7(3 - x)\text{ to both sides} \right) \\[6pt] \implies & \log_7\left( (1 - 2x)(3 - x) \right) & = & 1 & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] \implies & (1 - 2x)(3 - x) & = & 7^1 & \quad & \left( \text{rewriting in exponential form} \right) \\[6pt] \implies & 2x^2 - 7x + 3 & = & 7 & \quad & \left( \text{distributing} \right) \\[6pt] \implies & 2x^2 - 7x - 4 & = & 0 & \quad & \left( \text{subtracting }7\text{ from both sides} \right) \\[6pt] \implies & (2x + 1)(x - 4) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt] \end{array} \nonumber \]Therefore, either \( x = -\frac{1}{2} \) or \( x = 4 \). We now check our candidate solutions (as we always must do when solving logarithmic equations).
Substituting \( x = -\frac{1}{2} \) into the original equation, we get\[ \begin{array}{rrclcl} & \log_7\left( 1 - 2\left( -\dfrac{1}{2} \right) \right) & \overset{\text{?}}{=} & 1 - \log_7\left( 3 - \left( -\dfrac{1}{2} \right) \right) & \quad & \left( \text{substituting} \right) \\[6pt] \implies & \log_7\left( 1 + 1 \right) & \overset{\text{?}}{=} & 1 - \log_7\left( \dfrac{7}{2} \right) & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & \log_7\left( 2 \right) & \overset{\text{?}}{=} & 1 - \left( \log_7( 7) - \log_7(2) \right) & \quad & \left( \text{simplifying and Laws of Logarithms} \right) \\[6pt] \implies & \log_7\left( 2 \right) & \overset{\text{?}}{=} & 1 - \log_7( 7) + \log_7(2) & \quad & \left( \text{distributing} \right) \\[6pt] \implies & \log_7\left( 2 \right) & \overset{\text{?}}{=} & 1 - 1 + \log_7(2) & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] \implies & \log_7\left( 2 \right) & \overset{\checkmark}{=} & \log_7(2) & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]Thus, \( x = -\frac{1}{2} \) is a solution.
Substituting \( x = 4 \) into the original equation, we get the following.\[ \begin{array}{rrcl} & \log_7(1 - 2(4)) & \overset{\text{?}}{=} & 1 - \log_7(3 - 4) \\[6pt] \implies & \log_7(-7) & \overset{\text{?}}{=} & \log_7(-1) \\[6pt] \end{array} \nonumber \]We can stop at this point and confidently state that \( x = 4 \) is an extraneous solution.
WARNING: Our determination that \( x = 4 \) is extraneous is not because \( -7 \neq -1 \). Instead, it is due to the fact that the argument of a logarithm can never be negative!
Hence, the only solution to the equation is \( -\frac{1}{2} \).
Solve \( \ln x=3\) and use a graph to interpret the solution.
- Solution
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\[ \ln x = 3 \implies x = e^3. \nonumber \]Figure \( \PageIndex{ 3 } \) represents the graph of the equation. On the graph, the \( x \)-coordinate of the point at which the two graphs intersect is close to \( 20 \). In other words \(e^3 \approx 20\). A calculator gives a better approximation: \(e^3 \approx 20.0855\).
Figure \( \PageIndex{ 3 } \): The graphs of \(y = \ln x\) and \(y=3\) cross at the point \( (e^3 ,3)\), which is approximately \( (20.0855, 3) \).
Use graphing technology to estimate the approximate solution to the logarithmic equation \(2^x =1000\) to \( 2 \) decimal places.
Using the One-to-One Property of Logarithms to Solve Logarithmic Equations
As with exponential equations, we can use the One-to-One Property to solve logarithmic equations. The One-to-One Property of logarithmic functions tells us that, for any real numbers \(x>0\), \(S>0\), \(T>0\) and any positive real number \(b\), where \(b \neq 1\),\[ \log_b S = \log_b T \text{ if and only if } S=T. \nonumber \]For example,\[\text{If } \log_2 (x−1)= \log_2 (8), \text{ then } x−1 = 8. \nonumber \]So, if \(x−1=8\), then we can solve for \(x\), and we get \(x=9\). To check, we can substitute \(x=9\) into the original equation: \( \log_2 ( 9−1 )= \log_2 ( 8 )=3\). In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use the Laws of Logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.
For example, consider the equation \( \log( 3x−2 ) − \log( 2 ) = \log( x+4 )\). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the One-to-One Property to solve for \(x\):\[ \begin{array}{rrclcl} & \log(3x - 2) - \log(2) & = & \log(x + 4) & & \\[6pt] \implies & \log\left( \dfrac{3x - 2}{2} \right) & = & \log(x + 4) & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt] \implies & \dfrac{3x - 2}{2} & = & x + 4 & \quad & \left( \text{One-to-One Property} \right) \\[6pt] \implies & 3x - 2 & = & 2x + 8 & \quad & \left( \text{multiplying both sides by }2 \right) \\[6pt] \implies & x & = & 10 & \quad & \left( \text{subtracting }2x\text{ and adding }2\text{ to both sides} \right) \\[6pt] \end{array} \nonumber \]To check the result, substitute \(x=10\) into the original equation.\[ \begin{array}{rrcl} & \log\left( 3(10) - 2 \right) - \log(2) & \overset{\text{?}}{=} & \log\left( 10 + 4 \right) \\[6pt] \implies & \log\left( 28 \right) - \log(2) & \overset{\text{?}}{=} & \log\left( 14 \right) \\[6pt] \implies & \log\left( \dfrac{28}{2} \right) & \overset{\text{?}}{=} & \log\left( 14 \right) \\[6pt] \implies & \log\left( 14 \right) & \overset{\checkmark}{=} & \log\left( 14 \right) \\[6pt] \end{array} \nonumber \]Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.
Solve \(\ln( x^2 ) = \ln(2x+3)\).
- Solution
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\[ \begin{array}{rrclcl} & \ln(x^2) & = & \ln(2x + 3) & & \\[6pt] \implies & x^2 & = & 2x + 3 & \quad & \left( \text{One-to-One Property} \right) \\[6pt] \implies & x^2 - 2x - 3 & = & 0 & \quad & \left( \text{subtracting }2x \text{ and }3\text{ from both sides} \right) \\[6pt] \implies & (x - 3)(x + 1) & = & 0 & \quad & \left( \text{factoring the trinomial} \right) \\[6pt] \end{array} \nonumber \]Therefore, \( x = 3\) or \( x = −1\)
We leave it to the reader to check the solutions; however, it is important to note that, while the solution \(−1\) is negative, it checks when substituted into the original equation because the argument of the logarithm functions is still positive.
Solve \(\ln( x^2 ) = \ln 1\).
Access these online resources for additional instruction and practice with exponential and logarithmic equations.