Pre-Calculus: Solving Exponential Equations

Exponential equations, where the variable resides in an exponent, model phenomena from compound interest and population growth to radioactive decay and sound intensity. Mastering the techniques to solve them unlocks your ability to analyze these real-world systems quantitatively.

The Foundational Strategy: Isolating the Exponential

The universal first step in solving any exponential equation is to isolate the exponential expression on one side of the equation. An exponential expression is any term of the form $b^{f(x)}$, where $b$ is the base and $f(x)$ is the exponent containing the variable. Consider the equation $5\cdot 3^{2x-1}+7=22$. Your immediate goal is to get $3^{2x-1}$ by itself.

1. Subtract 7 from both sides: $5\cdot 3^{2x-1}=15$.
2. Divide both sides by 5: $3^{2x-1}=3$.

Only once the exponential term is isolated can you effectively choose your solving tactic. The choice hinges on a critical observation: can you express both sides of the equation with the same base?

Case 1: Solving by Recognizing a Common Base

If both sides of the isolated equation can be written as powers of the same base, you can solve using a simple but powerful property: if $b^M=b^N$ and $b$ is positive and not equal to 1, then $M=N$. This works because exponential functions are one-to-one.

In our example, we isolated $3^{2x-1}=3$. Notice that the right side, 3, can be written as $3^1$. This gives us $3^{2x-1}=3^1$. Since the bases are identical (and positive and not 1), the exponents must be equal: $$2x-1=1$$ Solving this linear equation gives $2x=2$, so $x=1$. Always verify your solution by substituting it back into the original equation: $5\cdot 3^{2(1)-1}+7=5\cdot 3^1+7=15+7=22$. It checks out.

Example with Manipulation: Solve $8^x=32$. While 8 and 32 are different, they are both powers of 2: $8=2^3$ and $32=2^5$. Rewrite the equation: $(2^3{)}^x=2^5$. Apply the power of a power rule: $2^{3x}=2^5$. Now the bases are the same, so $3x=5$, yielding $x=\frac{5}{3}$.

Case 2: Solving by Applying Logarithms

When you cannot rewrite both sides with a common base (e.g., $7^x=123$ or $e^{2x}=5$), you must employ logarithms. A logarithm is the inverse function of an exponent. The defining principle you will use is the logarithm of a power property: ${\log }_b(M^p)=p\cdot {\log }_b(M)$. This property allows you to "bring the exponent down" in front of the log, turning an exponential equation into a manageable algebraic one.

The general procedure is:

1. Isolate the exponential term.
2. Take the logarithm of both sides of the equation. You may use a common logarithm ($\log$, base 10) or a natural logarithm ($\ln$, base $e$). The choice is often based on the original base; use $\ln$ when the base is $e$.
3. Apply the power rule to bring the variable exponent out in front.
4. Solve the resulting equation for the variable.

Step-by-Step Example with Common Log: Solve $5^{x+2}=30$. The exponential is already isolated. Taking the common log of both sides is a reliable choice: $$\log (5^{x+2})=\log (30)$$ Apply the power rule: $(x+2)\log (5)=\log (30)$. Now solve for $x$ as if $\log (5)$ and $\log (30)$ were just numbers: $$x+2=\frac{\log (30)}{\log (5)}$$ $$x=\frac{\log (30)}{\log (5)}-2$$ Use your calculator to approximate: $x\approx \frac{1.4771}{0.6990}-2\approx 2.113-2=0.113$.

Step-by-Step Example with Natural Log: Solve $3e^{4t}=90$. First, isolate the exponential: $e^{4t}=30$. Since the base is $e$, using the natural logarithm $\ln$ is most efficient: $$\ln (e^{4t})=\ln (30)$$ A key property is that $\ln (e^u)=u$. Applying this gives: $$4t=\ln (30)$$ $$t=\frac{\ln (30)}{4}$$ This is the exact answer. The approximate value is $t\approx \frac{3.4012}{4}=0.8503$.

Advanced Application: Equations in Quadratic Form

Some exponential equations can be transformed into a familiar quadratic form using substitution. Look for an equation like $e^{2x}-5e^x+6=0$. Notice that $e^{2x}=(e^x{)}^2$.

Let $u=e^x$. Then $e^{2x}=u^2$. Substitute into the original equation: $$u^2-5u+6=0$$ This factors as $(u-2)(u-3)=0$, so $u=2$ or $u=3$. Remember, $u=e^x$. Therefore, $e^x=2$ or $e^x=3$. Now solve each simple exponential equation using $\ln$: $x=\ln (2)$ or $x=\ln (3)$. Always verify that these solutions are in the domain, which they are since $e^x$ is always positive.

Common Pitfalls

1. Misapplying Logarithm Rules: The log of a sum is not the sum of the logs. For example, $\ln (x+2)$ cannot be simplified to $\ln (x)+\ln (2)$. You may only apply the product, quotient, or power rules when the entire argument of the log is a product, quotient, or power. A good habit is to isolate the exponential term before taking logs.

1. Forgetting the "Exact" or "Approximate" Distinction: In many contexts, an answer like $x=\frac{\ln (7)}{3}$ is the preferred exact form. It is more precise than a decimal approximation like 0.648. Unless a problem specifically asks for a numerical approximation, leave your answer in exact logarithmic form.

1. Ignoring Domain Restrictions: The argument of any logarithm must be positive. If your algebraic manipulation leads to an equation like ${\log }_2(x-5)=3$, you must remember that $x-5>0$, so $x>5$. If your calculated solution does not satisfy this, it is extraneous and must be discarded. Similarly, bases of exponential expressions must be positive real numbers not equal to 1.

1. Dropping Coefficients When Taking Logs: When you take the log of both sides of $5\cdot 3^x=20$, write it as $\log (5\cdot 3^x)=\log (20)$. You can then use the product rule to write $\log (5)+\log (3^x)=\log (20)$, which becomes $\log (5)+x\log (3)=\log (20)$. Do not mistakenly write $5\cdot \log (3^x)=\log (20)$.

Summary

• The core strategy begins with isolating the exponential expression $b^{f(x)}$ on one side of the equation.
• If both sides can be written as powers of the same base, use the property $b^M=b^N⟹M=N$ to equate exponents and solve.
• If a common base is not possible, take the logarithm of both sides. Use the power rule ${\log }_b(M^p)=p{\log }_b(M)$ to bring the variable exponent down in front, converting the problem into a linear or algebraic equation.
• You can use either common logarithms ($\log$) or natural logarithms ($\ln$); $\ln$ is particularly efficient when the exponential base is $e$.
• Some equations can be solved by recognizing a quadratic form after a clever substitution like $u=e^x$.
• Always be mindful of domain restrictions for logarithms and verify your solutions in the context of the original problem.