Table of Contents
In this chapter, we focus on how to use logarithms to solve equations. You will rely heavily on the basic properties and laws of logarithms, which are treated in the parent chapter, and on your knowledge of exponential functions.
Throughout, assume all logarithms are defined (their arguments are positive and bases are valid).
1. Strategy overview
When you see an equation involving logarithms or exponentials, the general goal is to simplify it into one of these two familiar forms:
- A pure exponential equation:
$$a^{f(x)} = b$$
which you solve by taking a logarithm of both sides. - A pure logarithmic equation:
$$\log_a(f(x)) = c$$
which you solve by rewriting in exponential form:
$$f(x) = a^c.$$
Most realistic equations take a few steps to turn into one of these clean forms.
Two basic moves are used again and again:
- Rewrite using log/exponential relationships and logarithm laws.
- Isolate the log or exponential expression before applying an inverse operation.
Always check for extraneous solutions by verifying that log arguments are positive.
2. Solving simple logarithmic equations
These are equations where a single logarithm equals a number.
2.1 Single log equals a constant
Typical form:
$$\log_b(f(x)) = k.$$
Step pattern:
- Ensure the log is by itself on one side.
- Rewrite using the definition of logarithm:
$$\log_b(f(x)) = k \quad \Longleftrightarrow \quad f(x) = b^k.$$ - Solve the resulting (usually algebraic) equation.
- Check that $f(x) > 0$ for your solution(s).
Example:
Solve $\log_2(x - 3) = 4$.
- Already isolated.
- Rewrite:
$$x - 3 = 2^4 = 16.$$ - Solve:
$$x = 19.$$ - Check domain: $x - 3 > 0 \Rightarrow x > 3$. The solution $x=19$ is valid.
2.2 Natural and common logs
Nothing changes conceptually if the base is $e$ or $10$:
- $\ln(f(x)) = k \Rightarrow f(x) = e^k$.
- $\log(f(x)) = k \Rightarrow f(x) = 10^k$ (for common log base $10$).
Example:
Solve $\ln(2x + 1) = 3$.
Rewrite:
$$2x + 1 = e^3 \quad\Rightarrow\quad 2x = e^3 - 1 \quad\Rightarrow\quad x = \dfrac{e^3 - 1}{2}.$$
Check: $2x + 1 > 0$ is satisfied by this $x$, so it is a valid solution.
3. Solving exponential equations using logarithms
Here the variable is in an exponent, and you usually take logs on both sides.
3.1 Single exponential term
Typical form:
$$a^{f(x)} = c.$$
Strategy:
- Isolate the exponential expression if necessary.
- Take a logarithm (any convenient base) of both sides.
- Use logarithm laws to bring the exponent down.
- Solve the resulting algebraic equation.
Example:
Solve $3^{2x - 1} = 7$.
- Exponential part already isolated.
- Take natural logs of both sides:
$$\ln(3^{2x - 1}) = \ln 7.$$ - Bring down the exponent:
$$(2x - 1)\ln 3 = \ln 7.$$ - Solve for $x$:
$x - 1 = \frac{\ln 7}{\ln 3}, \quad 2x = 1 + \frac{\ln 7}{\ln 3},$$
$$x = \frac{1}{2} + \frac{1}{2}\cdot\frac{\ln 7}{\ln 3}.$$
Any log base would work; using $\ln$ or $\log$ is just convention.
3.2 Exponential equations with different bases
If the bases cannot be rewritten to match, logarithms are practically required.
Example:
Solve $5^x = 2^x \cdot 10$.
- Isolate terms with exponentials on one side if possible:
$$\frac{5^x}{2^x} = 10.$$ - Combine using exponent laws:
$$\left(\frac{5}{2}\right)^x = 10.$$ - Take logs (say natural logs) on both sides:
$$\ln\left(\left(\frac{5}{2}\right)^x\right) = \ln 10.$$ - Bring down the exponent:
$$x \ln\left(\frac{5}{2}\right) = \ln 10.$$ - Solve:
$$x = \frac{\ln 10}{\ln(5/2)}.$$
No simpler exact form is needed; this is a perfectly valid algebraic solution.
4. Equations with several logarithms
These involve sums, differences, or multiple log terms. The laws of logarithms allow you to combine or expand them to reach a solvable form.
4.1 Combining logs into a single logarithm
Typical target:
$$\log_b(A) + \log_b(B) = \log_b(AB),$$
$$\log_b(A) - \log_b(B) = \log_b\left(\dfrac{A}{B}\right).$$
Strategy:
- Combine multiple log terms on one side into a single logarithm.
- Rewrite the resulting equation in exponential form.
- Solve the resulting algebraic equation.
- Check that all original log arguments are positive.
Example:
Solve $\log_3(x) + \log_3(x - 2) = 2$.
- Combine:
$$\log_3(x) + \log_3(x - 2) = \log_3\bigl(x(x - 2)\bigr).$$
So the equation becomes
$$\log_3\bigl(x(x - 2)\bigr) = 2.$$ - Rewrite in exponential form:
$$x(x - 2) = 3^2 = 9.$$ - Solve the quadratic:
$$x^2 - 2x - 9 = 0.$$
Use the quadratic formula:
$$x = \frac{2 \pm \sqrt{4 + 36}}{2}
= \frac{2 \pm \sqrt{40}}{2}
= 1 \pm \sqrt{10}.$$ - Check domain restrictions: both arguments must be positive:
- For $\log_3(x)$: $x > 0$.
- For $\log_3(x - 2)$: $x - 2 > 0 \Rightarrow x > 2$.
So $x$ must be $> 2$. Among our candidates:
- $x = 1 - \sqrt{10}$ is negative, so discard.
- $x = 1 + \sqrt{10} > 2$, so keep.
Final solution: $x = 1 + \sqrt{10}$.
4.2 Moving coefficients using the power rule
If an equation contains a constant multiple of a logarithm, use the power rule:
$$c \log_b(A) = \log_b(A^c).$$
This helps you combine multiple terms.
Example:
Solve $2\log_5(x) - \log_5(3) = 1$.
- Use the power rule:
$\log_5(x) = \log_5(x^2).$$
So the equation becomes
$$\log_5(x^2) - \log_5(3) = 1.$$ - Combine using the quotient rule:
$$\log_5\left(\dfrac{x^2}{3}\right) = 1.$$ - Rewrite in exponential form:
$$\dfrac{x^2}{3} = 5^1 = 5.$$ - Solve:
$$x^2 = 15 \quad\Rightarrow\quad x = \pm\sqrt{15}.$$ - Domain check: original equation includes $\log_5(x)$, so $x > 0$.
Discard $x = -\sqrt{15}$, keep $x = \sqrt{15}$.
5. Solving equations with logs on both sides
Here you aim to isolate a single logarithm (or same-base logs) on each side and then set their arguments equal, or rewrite both sides and equate exponents.
5.1 Equal logs: equate arguments
If you have
$$\log_b(A) = \log_b(B)$$
with the same base $b$ and both sides defined, then
$$A = B.$$
Example:
Solve $\log_4(3x - 1) = \log_4(2x + 7)$.
- Since the logs have the same base and are equal, equate arguments:
$x - 1 = 2x + 7.$$ - Solve:
$x - 2x = 7 + 1 \Rightarrow x = 8.$$ - Domain check:
- $3x - 1 > 0 \Rightarrow 3(8) - 1 = 23 > 0.$
- $2x + 7 > 0 \Rightarrow 2(8) + 7 = 23 > 0.$
Both valid, so $x = 8$ is the solution.
5.2 Multiple terms before equating arguments
Often you must first combine terms so that each side is a single logarithm.
Example:
Solve $\log_2(x) + \log_2(x - 1) = \log_2(6)$.
- Combine the logs on the left:
$$\log_2(x) + \log_2(x - 1) = \log_2\bigl(x(x - 1)\bigr).$$
So:
$$\log_2\bigl(x(x - 1)\bigr) = \log_2(6).$$ - Equate arguments:
$$x(x - 1) = 6.$$ - Solve:
$$x^2 - x - 6 = 0,$$
so
$$x = \frac{1 \pm \sqrt{1 + 24}}{2}
= \frac{1 \pm 5}{2}
\Rightarrow x = 3 \text{ or } x = -2.$$ - Domain check for logs:
- $x > 0$ and $x - 1 > 0 \Rightarrow x > 1$.
From the candidates, only $x = 3$ fits.
Final answer: $x = 3$.
6. Mixed exponential–logarithmic equations
Sometimes variables appear in both logs and exponentials; often you cannot combine them into a single log or a single exponential. Many of these can be tackled algebraically, but some require numerical/graphical methods.
6.1 Rearranging to use log/exponential inverses
Look for ways to:
- Turn an exponential expression into a log equation, or
- Turn a log equation into an exponential equation,
by isolating part of the expression.
Example (algebraically solvable):
Solve $2^x = x$.
This equation does not have a simple algebraic solution. Typical approaches:
- Use a graphing method (graph $y = 2^x$ and $y = x$).
- Use numerical methods (e.g. trial-and-improvement, or more advanced methods in later courses).
So not every equation involving exponentials or logarithms can be solved exactly using algebra alone. In this chapter we mainly focus on those that can.
6.2 Equations solvable by taking logs after algebraic manipulation
Example:
Solve $3^{x+1} = 5x$.
- Rewrite exponential:
$^{x+1} = 3\cdot 3^x,$$
so the equation becomes
$\cdot 3^x = 5x.$$ - Here $x$ appears both in an exponent and as a factor. There is no straightforward way to isolate $x$ with just algebraic/logarithmic manipulations.
- This is another example where numerical or graphical methods are needed.
The key point: knowing when an algebraic-logarithmic solution is not available is part of solving equations involving logs.
7. Domain restrictions and extraneous solutions
Equations with logarithms require extra attention to the domain. Every solution candidate must respect:
- Logarithm arguments must be positive:
$$\log_b(A) \text{ is defined only if } A > 0.$$ - Bases must be valid, but in pre-designed exercises the base is usually acceptable (no need to check unless you manipulate it yourself).
Whenever you solve a transformed version of the equation (e.g., after squaring, expanding, or using log laws), you might introduce solutions that do not satisfy the original equation. These are extraneous solutions.
7.1 Systematic checking
To avoid mistakes:
- List all explicit conditions from log arguments (like $x > 2$, $x \neq -1$, etc.).
- Solve the equation to get candidate solutions.
- Substitute each candidate back into the original equation to check:
- Are all log arguments positive?
- Does each side numerically match?
Example:
Solve $\log_2(x - 1) + \log_2(x - 3) = 3$.
- Domain conditions:
- $x - 1 > 0 \Rightarrow x > 1$,
- $x - 3 > 0 \Rightarrow x > 3$,
so overall $x > 3$. - Combine logs:
$$\log_2\bigl((x - 1)(x - 3)\bigr) = 3.$$ - Rewrite:
$$(x - 1)(x - 3) = 2^3 = 8.$$ - Expand:
$$x^2 - 4x + 3 = 8 \Rightarrow x^2 - 4x - 5 = 0.$$ - Solve:
$$x = \frac{4 \pm \sqrt{16 + 20}}{2}
= \frac{4 \pm \sqrt{36}}{2}
= \frac{4 \pm 6}{2}
\Rightarrow x = 5 \text{ or } x = -1.$$ - Apply domain $x > 3$. Discard $x = -1$, keep $x = 5$.
- Check in original equation:
- $\log_2(5 - 1) + \log_2(5 - 3) = \log_2(4) + \log_2(2) = 2 + 1 = 3$.
So $x = 5$ is valid.
8. Typical patterns to recognize
While the detailed algebra can vary, many log/exponential equations fit into a small number of templates. When you see:
- Single log equals number:
- Isolate log, rewrite as exponential.
- Single exponential equals number:
- Isolate exponential, take logs of both sides.
- Sum/difference of logs equals number:
- Combine logs into one using product/quotient/power rules, then rewrite as exponential.
- Log equals log (same base):
- Ensure each side is one log, then equate arguments.
- Exponentials with different bases:
- Try to combine, if not possible, take logs on both sides.
- Exponentials and linear/polynomial terms mixed (variable appears both in exponent and outside):
- Often not solvable in closed form; consider numerical/graphical approaches.
Understanding which pattern you see will guide the steps you take.
9. Practice structure
When you practice solving logarithmic and exponential equations, for each problem:
- Identify the type (from the patterns above).
- Plan: decide whether you will:
- combine logs,
- rewrite as exponential,
- take logs on both sides,
- or accept a numerical/graphical solution.
- Execute the algebra carefully, using laws of exponents and logs.
- Check:
- Apply domain restrictions from log arguments.
- Substitute into the original equation to confirm.
Consistently following these steps will make solving logarithmic and exponential equations systematic rather than guesswork.