Solving Exponential and Logarithmic Equations

You have been solving equations your entire math career. Whether it was figuring out what number plus 3 gives you 7, or finding which value of $x$ makes $2x - 5 = 11$ true, the goal has always been the same: undo the operations to isolate what you do not know. Exponential and logarithmic equations are no different. The variable is just hiding in a new place (the exponent), and you need a new tool (logarithms) to coax it out.

Think about it this way: if someone asked you “2 to what power equals 8?” you could probably figure out the answer is 3 without much trouble. But what if they asked “5 to what power equals 17?” Now the answer is not a nice integer, and you cannot just guess. This is where logarithms become essential. They are the mathematical tool designed specifically to answer the question “what exponent gives me this result?”

In this lesson, you will learn systematic methods for solving equations where the variable appears in an exponent or inside a logarithm. You will also discover why checking your answers is especially important when logarithms are involved, since some “solutions” turn out to be impostors.

Core Concepts

Solving Exponential Equations by Rewriting with the Same Base

The simplest exponential equations can be solved without logarithms at all. The key insight is the one-to-one property of exponential functions: if two powers of the same base are equal, then their exponents must be equal.

$$\text{If } b^x = b^y, \text{ then } x = y \quad (b > 0, b \neq 1)$$

This works because exponential functions never repeat values. If $2^3 = 8$ and $2^4 = 16$, there is no other exponent that gives you 8 or 16. Each output corresponds to exactly one input.

So when you see an equation like $4^x = 64$, your first instinct should be to ask: “Can I write both sides as powers of the same base?” Since $4 = 2^2$ and $64 = 2^6$, you can rewrite the equation as:

$$(2^2)^x = 2^6$$ $$2^{2x} = 2^6$$

Now the one-to-one property tells you that $2x = 6$, so $x = 3$.

This technique is limited to equations where both sides can be expressed as powers of a common base, typically 2, 3, 5, or 10. When that is not possible, you need logarithms.

Solving Exponential Equations Using Logarithms

When you cannot rewrite both sides with a common base, logarithms provide the solution. The strategy is straightforward: take the logarithm of both sides, then use the power rule for logarithms to bring the exponent down where you can work with it.

For an equation like $5^x = 17$:

Step 1: Take the logarithm of both sides. You can use any base (common log, natural log, or any other), but log base 10 or natural log are most convenient for calculator work.

$$\log(5^x) = \log(17)$$

Step 2: Apply the power rule ($\log(a^n) = n \cdot \log(a)$) to bring down the exponent:

$$x \cdot \log(5) = \log(17)$$

Step 3: Solve for $x$:

$$x = \frac{\log(17)}{\log(5)}$$

This is your exact answer. If you need a decimal approximation, use a calculator: $x \approx 1.76$.

Notice that the answer $\frac{\log(17)}{\log(5)}$ is also equal to $\log_5(17)$ by the change of base formula. This makes sense: “5 to what power equals 17?” is exactly what $\log_5(17)$ asks.

Solving Logarithmic Equations

Logarithmic equations come in several forms, and each has its own approach.

Type 1: Logarithm equals a number

When you have $\log_b(x) = c$, convert to exponential form: $x = b^c$.

For example, $\log_3(x) = 4$ becomes $x = 3^4 = 81$.

Type 2: Logarithm equals logarithm (same base)

Just as exponential functions are one-to-one, so are logarithmic functions. If $\log_b(A) = \log_b(B)$, then $A = B$.

For example, if $\log_2(x + 5) = \log_2(3x - 1)$, then $x + 5 = 3x - 1$, giving $x = 3$.

Type 3: Multiple logarithms that can be combined

Use logarithm properties to combine terms, then solve:

$$\log(x) + \log(x + 3) = 1$$ $$\log(x(x + 3)) = 1$$ $$x(x + 3) = 10^1 = 10$$ $$x^2 + 3x - 10 = 0$$

This factors to $(x + 5)(x - 2) = 0$, giving $x = -5$ or $x = 2$. But wait! You must check for extraneous solutions.

Checking for Extraneous Solutions

This is the most important warning in this entire lesson: logarithmic equations frequently produce extraneous solutions that do not satisfy the original equation.

Why does this happen? The domain of $\log_b(x)$ requires $x > 0$. When you use logarithm properties to combine or manipulate terms, you may introduce solutions that violate this requirement.

In the example above, $x = -5$ is extraneous because it would require evaluating $\log(-5)$, which is undefined. Only $x = 2$ is valid:

• Check: $\log(2) + \log(2 + 3) = \log(2) + \log(5) = \log(10) = 1$ ✓

Always substitute your solutions back into the original equation to verify they work.

Exponential Equations in Quadratic Form

Some exponential equations are disguised quadratics. The equation $e^{2x} - 5e^x + 6 = 0$ looks complicated, but if you let $u = e^x$, it becomes:

$$u^2 - 5u + 6 = 0$$

This factors as $(u - 2)(u - 3) = 0$, giving $u = 2$ or $u = 3$.

Now substitute back: $e^x = 2$ or $e^x = 3$.

Taking natural logs: $x = \ln(2)$ or $x = \ln(3)$.

Both solutions are valid because both 2 and 3 are positive (so $e^x$ can equal them).

Exponential and Logarithmic Inequalities

Solving inequalities requires careful attention to whether functions are increasing or decreasing.

Exponential inequalities (base $b > 1$): The exponential function is increasing, so the inequality direction is preserved.

$$3^x > 81$$ $$3^x > 3^4$$ $$x > 4$$

Exponential inequalities (base $0 < b < 1$): The exponential function is decreasing, so the inequality direction reverses.

$$\left(\frac{1}{2}\right)^x > 8$$ $$\left(\frac{1}{2}\right)^x > \left(\frac{1}{2}\right)^{-3}$$ $$x < -3$$

Logarithmic inequalities: Since $\log_b(x)$ is increasing when $b > 1$, the direction is preserved (but remember $x$ must be positive):

$$\log_2(x - 1) > 3$$ $$x - 1 > 2^3 = 8$$ $$x > 9$$

Applications: Finding Time, Rate, or Initial Value

Exponential and logarithmic equations appear constantly in applications. The general exponential model is:

$$A(t) = A_0 \cdot b^{t/k}$$

where:

• $A(t)$ is the amount at time $t$
• $A_0$ is the initial amount
• $b$ is the growth/decay factor
• $k$ is the time for one growth cycle

Different problems ask you to solve for different unknowns:

• Finding time: Given initial and final amounts, solve for $t$ using logarithms
• Finding rate: Given initial amount and amount after known time, solve for $b$ or $k$
• Finding initial value: Given current amount and time elapsed, solve for $A_0$

Newton’s Law of Cooling

When a hot object is placed in a cooler environment (or vice versa), its temperature changes according to:

$$T(t) = T_s + (T_0 - T_s)e^{-kt}$$

where:

• $T(t)$ is the temperature at time $t$
• $T_s$ is the surrounding (ambient) temperature
• $T_0$ is the initial temperature of the object
• $k$ is a positive cooling constant (depends on the object and environment)

This model says that temperature change is proportional to the difference between the object and its surroundings. A hot cup of coffee cools quickly at first (when the temperature difference is large) and more slowly as it approaches room temperature.

Logistic Growth Models

Not everything grows exponentially forever. Populations, for instance, eventually hit limits due to resources, space, or competition. The logistic model captures this:

$$P(t) = \frac{L}{1 + Ce^{-kt}}$$

where:

• $L$ is the carrying capacity (maximum sustainable population)
• $C$ is a constant determined by initial conditions
• $k$ is the growth rate

The logistic curve starts with near-exponential growth, then slows as it approaches the carrying capacity $L$. Solving logistic equations often requires isolating the exponential term and taking logarithms.

Notation and Terminology

Term	Meaning	Example
One-to-one property	If $b^x = b^y$, then $x = y$	$2^x = 2^5 \Rightarrow x = 5$
Taking the log	Applying logarithm to both sides	$5^x = 12 \Rightarrow x = \frac{\log 12}{\log 5}$
Extraneous solution	Answer that does not satisfy original equation	Common in log equations due to domain restrictions
Carrying capacity	Maximum sustainable value in logistic growth	The $L$ in $P(t) = \frac{L}{1 + Ce^{-kt}}$
Cooling constant	Rate at which temperature approaches ambient	The $k$ in Newton’s Law of Cooling

Examples

Key Properties and Rules

Strategies for Solving Exponential Equations

Situation	Strategy	Example
Both sides can be same base	Use one-to-one property	$8^x = 32 \Rightarrow 2^{3x} = 2^5 \Rightarrow x = \frac{5}{3}$
Bases cannot match	Take log of both sides	$3^x = 20 \Rightarrow x = \frac{\log 20}{\log 3}$
Quadratic form	Substitute $u = b^x$	$4^x - 6 \cdot 2^x + 8 = 0 \Rightarrow u^2 - 6u + 8 = 0$

Strategies for Solving Logarithmic Equations

Situation	Strategy	Example
Single log = number	Convert to exponential form	$\log_2(x) = 5 \Rightarrow x = 2^5 = 32$
Log = log (same base)	Set arguments equal	$\log(2x) = \log(x + 5) \Rightarrow 2x = x + 5$
Multiple logs	Combine using properties	$\log x + \log y = \log(xy)$

Critical Reminders

1. Always check for extraneous solutions in logarithmic equations
2. The argument of any logarithm must be positive
3. When taking logs, you can use any base (the solution will be the same)
4. For inequalities with decreasing functions, reverse the inequality sign

Real-World Applications

Finding Doubling Time and Half-Life

Doubling time is how long it takes a quantity to double. For exponential growth $A(t) = A_0 e^{kt}$:

$$2A_0 = A_0 e^{kt_d}$$ $$2 = e^{kt_d}$$ $$t_d = \frac{\ln 2}{k}$$

Half-life is how long it takes a quantity to halve (common in radioactive decay):

$$t_{1/2} = \frac{\ln 2}{k}$$

Carbon Dating

Living organisms contain carbon-14, which decays with a half-life of about 5,730 years. Scientists measure the remaining carbon-14 to estimate age:

$$A(t) = A_0 \cdot \left(\frac{1}{2}\right)^{t/5730}$$

If an artifact has 35% of its original carbon-14, solving $0.35 = (0.5)^{t/5730}$ tells us its age.

Population Projections

City planners, ecologists, and economists all use exponential and logistic models to forecast population:

• Short-term growth often follows exponential models
• Long-term growth typically follows logistic models as resources become limited

Cooling and Heating Problems

Newton’s Law of Cooling applies to:

• Forensic science (determining time of death)
• Food safety (how long until food reaches safe temperature)
• Engineering (cooling electronic components)

Investment Planning

The compound interest formula $A = P(1 + r/n)^{nt}$ requires logarithms to solve for time:

“How long until my investment doubles at 7% interest?” requires solving $2P = P(1.07)^t$.

Self-Test Problems

Summary

• One-to-one property: If $b^x = b^y$, then $x = y$. Use this when both sides can be written with the same base.
• Taking logarithms: When bases cannot match, take the log of both sides and use the power rule to bring down the exponent: $b^x = c \Rightarrow x = \frac{\log c}{\log b}$.
• Logarithmic equations: Convert to exponential form or use properties to combine logs, then solve.
• Extraneous solutions: Always check your answers in the original equation. Solutions that make any logarithm argument negative or zero are extraneous.
• Quadratic form: Equations like $e^{2x} - 5e^x + 6 = 0$ can be solved by substituting $u = e^x$.
• Inequalities: Preserve inequality direction for increasing functions ($b > 1$); reverse for decreasing functions ($0 < b < 1$).
• Newton’s Law of Cooling: $T(t) = T_s + (T_0 - T_s)e^{-kt}$ models how temperature approaches ambient temperature over time.
• Logistic growth: $P(t) = \frac{L}{1 + Ce^{-kt}}$ models growth that levels off at a carrying capacity $L$.
• Applications require setting up models, then using logarithms to solve for time, rate, or initial values.