8.3 Exponential and Logarithmic Equations

Exponential and logarithmic equations let you solve for unknowns that appear in exponents or inside logarithms. Mastering these techniques is essential because they show up constantly in modeling growth, decay, compound interest, and scaling problems. This section covers how to move between exponential and logarithmic forms, how to combine log expressions, and how to check that your answers are actually valid.

Solving exponential equations with logarithms

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Using logarithms to solve exponential equations

An exponential equation has the variable in the exponent: $a^x = b$, where $a$ is the base, $x$ is the unknown, and $b$ is a constant.

The core idea: logarithms are the inverse of exponentiation, so applying a log "brings down" the exponent and lets you isolate $x$.

Steps to solve $a^x = b$:

1. Confirm the equation is solvable: the base must satisfy $a > 0$ and $a \neq 1$, and the right side must be positive ($b > 0$). If $b \leq 0$, there's no solution because an exponential function never outputs zero or a negative number.
2. Take the log base $a$ of both sides: $\log_a(a^x) = \log_a(b)$
3. The left side simplifies because log and exponentiation cancel: $x = \log_a(b)$
4. If you need a decimal answer and your calculator only has ln or log, use the change of base formula: $x = \frac{\ln(b)}{\ln(a)}$ or $x = \frac{\log(b)}{\log(a)}$
5. Check by substituting back into the original equation.

Logarithms with different bases

When the base is $e$ (Euler's number $\approx 2.718$), use the natural logarithm (ln):

When the base is 10, use the common logarithm (log):

For any other base, you can use whichever log you want (ln or log) combined with the change of base formula. The key is consistency: take the same type of log on both sides.

Solving logarithmic equations with exponentiation

Using exponentiation to solve logarithmic equations

A logarithmic equation has the variable inside the log: $\log_a(x) = b$, where $a$ is the base, $x$ is the unknown argument, and $b$ is a constant.

Since exponentiation is the inverse of a logarithm, you "undo" the log by raising the base to both sides.

Steps to solve $\log_a(x) = b$:

1. Confirm the base satisfies $a > 0$ and $a \neq 1$.
2. Exponentiate both sides with base $a$: $a^{\log_a(x)} = a^b$
3. The left side simplifies: $x = a^b$
4. Check the domain: the argument of any logarithm must be strictly positive. Substitute your answer back into the original equation and verify that every log argument is positive. If it isn't, that solution is extraneous and must be rejected.

Special cases with natural and common logarithms

For natural log equations (base $e$):

For common log equations (base 10):

Always substitute back to verify. This is especially important when the original equation has the variable in multiple places, because extraneous solutions can sneak in.

Solving exponential and logarithmic equations

Condensing and expanding logarithms

Many equations have multiple log terms that need to be combined before you can solve. The three properties you'll use constantly:

• Product rule: $\log_a(x) + \log_a(y) = \log_a(xy)$
• Quotient rule: $\log_a(x) - \log_a(y) = \log_a\left(\frac{x}{y}\right)$
• Power rule: $n \cdot \log_a(x) = \log_a(x^n)$

These only work when the logs have the same base. If they don't, you'll need the change of base formula first.

Example: Suppose you have $\log(x) + \log(x + 3) = 1$. Condense the left side using the product rule:

1. $\log(x(x + 3)) = 1$

2. Exponentiate: $x(x + 3) = 10^1 = 10$

3. Expand: $x^2 + 3x - 10 = 0$

4. Factor: $(x + 5)(x - 2) = 0$, so $x = -5$ or $x = 2$

5. Check domain: $x = -5$ fails because $\log(-5)$ is undefined. The only valid solution is $x = 2$.

Isolating exponential terms

When an equation mixes exponential and non-exponential terms, isolate the exponential part first using algebra (add, subtract, divide), then take a log of both sides.

Example: Solve $3 \cdot 2^x + 5 = 29$

1. Subtract 5: $3 \cdot 2^x = 24$
2. Divide by 3: $2^x = 8$
3. Take $\log_2$ of both sides: $x = \log_2(8) = 3$

If you end up with exponential terms on both sides (like $3^x = 5^{x-1}$), take ln or log of both sides, use the power rule to bring the exponents down, and then solve the resulting linear equation for $x$.

Domain and range of exponential and logarithmic functions

Understanding domain and range isn't just theory; it tells you which solutions are valid and which are extraneous.

Exponential function domain and range

For $f(x) = a^x$ where $a > 0$ and $a \neq 1$:

• Domain: all real numbers ($x \in \mathbb{R}$)
• Range: $(0, \infty)$, regardless of whether $a > 1$ or $0 < a < 1$. The output is always positive.

This is why $a^x = b$ has no solution when $b \leq 0$. An exponential function never reaches zero or goes negative.

Logarithmic function domain and range

For $g(x) = \log_a(x)$ where $a > 0$ and $a \neq 1$:

• Domain: $(0, \infty)$. You can only take the log of a positive number.
• Range: all real numbers ($(-\infty, \infty)$). A log can output any value, positive or negative.

This is why you must always check that the argument of every logarithm in your original equation is positive after substituting your answer.

Checking domain and range in context

When you solve an equation, apply domain restrictions from the original equation, not the simplified version. Algebraic manipulation can introduce extraneous solutions that satisfy the simplified form but violate the original domain.

In application problems, context adds further restrictions. If $x$ represents time, population, or a physical measurement, negative or zero values are typically meaningless. Always interpret your answer: does it make sense for the situation described?