Mathematics › 🌌 Grade 11 › Solving Exponential & Logarithmic Equations

Solving Exponential & Logarithmic Equations

Logarithms and exponentials are inverses — so an equation with the variable trapped in an exponent, or buried inside a log, can always be freed by applying the opposite operation to both sides.

Grade 11Algebra 2 / Pre-Calculus

Solving Exponential & Logarithmic Equations — illustration

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The big idea: Every log law from the previous lesson exists to be used, and solving equations is where that payoff arrives. An exponential equation bx = k is unlocked by taking a logarithm of both sides; a logarithmic equation is unlocked by exponentiating both sides. The one extra step logs demand that other algebra does not: a logarithm's argument must be positive, so every candidate solution has to be checked against that domain before it is accepted.

🎯 By the end, you'll be able to

Solve bˣ = k for x using logarithms and the change-of-base formula
Solve equations of the form a·bˣ = k by isolating the exponential first
Solve logarithmic equations by exponentiating and identify extraneous solutions
Apply exponential and logarithmic equations to compound interest, half-life, and scale problems

📎 You should already know

Logarithmic Slide Rule: Log Properties and Computations
Laws of exponents

Undoing an exponent with a logarithm

The previous lesson established that log_b and raising-to-the-b are inverse operations: they cancel. That is exactly what is needed when x is stuck in an exponent. Starting from b^x = k, take log_b of both sides and the left side collapses straight to x.

\[ b^{x} = k \iff x = \log_b k \]

An exponential equation and a logarithmic equation are two views of the same fact.

🔑 Any base of log will do

You rarely have a log_b button on a calculator. Instead take any logarithm of both sides — usually ln or log₀ — and use the power law to bring the exponent down: ln(b^x) = x·ln b. Then x = ln k / ln b, which is exactly the change-of-base formula in action.

Isolate the exponential first

When the exponential is not alone, treat it like any other algebraic quantity: get a·b^x = k down to b^x by itself — divide by a — before taking a log. Taking a log too early, while other terms are still attached, does not simplify anything because a log cannot be distributed over addition or multiplication by a constant that is outside the power.

🎮 Where the Curve Meets the Target LIVE

Draw the horizontal target line y = k against an exponential curve — the intersection point is the equation's solution.

✨ Reading the solution off a graph — and why k ≤ 0 fails

Graphically, solving b^x = k means asking where the curve y = b^x crosses the horizontal line y = k. Because an exponential curve is strictly increasing (or decreasing) and never flattens or turns back, that line crosses it in at most one place. And because b^x is always positive for b > 0, the curve never touches or crosses the x-axis — so the line only meets it when k > 0. If k is zero or negative, b^x = k has no real solution at all, and no amount of algebra will produce one.

Solving a logarithmic equation

When x is trapped inside a log instead of an exponent, do the reverse: exponentiate both sides using the log's base to strip the log away. First combine any sum or difference of logs into a single log using the product and quotient laws, then exponentiate.

\[ \log_b(x) = k \iff x = b^{k} \]

Exponentiating both sides with base b removes the logarithm.

⚠️ Check every solution against the domain

A logarithm is only defined for a positive argument. Solving log(x) + log(x − 3) = 1 combines to log(x(x−3)) = 1, giving x(x−3) = 10, i.e. x² − 3x − 10 = 0, which factors to x = 5 or x = −2. But x = −2 makes log(x) and log(x−3) undefined, so it must be rejected as extraneous — only x = 5 is a valid solution. Always substitute every candidate back into the original logs before accepting it. And remember, as in the previous lesson, that log(a + b) ≠ log a + log b — only a product splits into a sum of logs, so a sum inside a single log cannot be broken apart at all.

📝 Worked example: Solve 4 · 3ˣ = 108 for x.

Isolate the exponential first by dividing by 4: \( 3^x = 27 \).
Take a log of both sides (or recognize 27 as a power of 3): \( \ln(3^x) = \ln 27 \Rightarrow x\ln 3 = \ln 27 \).
Solve for x: \( x = \dfrac{\ln 27}{\ln 3} = 3 \), matching \( 3^3 = 27 \) directly.

✓ x = 3.

📝 Worked example: Solve log₂(x) + log₂(x − 2) = 3, checking for extraneous solutions.

Combine with the product law: \( \log_2\big(x(x-2)\big) = 3 \).
Exponentiate base 2: \( x(x-2) = 2^3 = 8 \), so \( x^2 - 2x - 8 = 0 \), which factors as \( (x-4)(x+2)=0 \), giving \( x = 4 \) or \( x = -2 \).
Check the domain: \( x = -2 \) makes \( \log_2(x) = \log_2(-2) \) undefined, so it is rejected. \( x = 4 \) gives \( \log_2 4 + \log_2 2 = 2+1=3 \), which checks out.

✓ x = 4 only; x = −2 is extraneous.

Applications: growth, decay, and scales

Compound interest, A = P(1 + r/n)^nt, traps the time t in an exponent — solving for how long an investment takes to reach a target uses exactly this log technique. Radioactive and other decay similarly buries time in an exponent; to solve for the time until a substance reaches a given fraction remaining, as in half-life problems, logs are unavoidable rather than optional — this is the tool that was missing when time first needed solving for. pH and other logarithmic scales run the idea in the opposite direction: they are already logs of a quantity, so converting a scale reading back into the underlying quantity means exponentiating.

Check your understanding

1. What is the first algebraic step in solving 5·2ˣ = 40?

The exponential must be isolated before taking a log, since a log of a sum or a scaled term does not simplify usefully. Dividing by 5 gives 2ˣ = 8.

2. For b > 0, how many real solutions does bˣ = −5 have?

bˣ is always positive for b > 0, so it can never equal a negative number — there is no real solution.

3. Solving log(x) + log(x − 3) = 1 gives candidate solutions x = 5 and x = −2. Which is correct?

x = −2 makes log(x) and log(x−3) undefined (negative arguments), so it is extraneous; only x = 5 satisfies the original equation.

4. Which statement is TRUE?

Only the product law is valid: log(ab) = log a + log b. There is no law for the log of a sum or difference of terms.

5. On the functionplot sim with target mode, the solution to bˣ = k is found where...

The target line y = k intersects the exponential curve at exactly the x-value that solves bˣ = k.

✅ Key takeaways

bˣ = k is solved by taking a logarithm of both sides: x = log_b k = ln k / ln b.
Isolate the exponential term before taking a log — a log cannot be distributed over addition or a leading coefficient.
bˣ = k has exactly one solution when k > 0, and none when k ≤ 0, since bˣ is always positive.
Logarithmic equations are solved by combining logs into one and then exponentiating both sides.
Every candidate solution to a log equation must be checked: a log's argument must be strictly positive, or the root is extraneous.