DWG NO. 3.5 — Lesson 5 of 6

Solving Exponential and Logarithmic Equations

Unit 3: Exponential and Logarithmic Functions · ~25 min

Objective Solve equations with the variable in an exponent or inside a logarithm, and check for extraneous solutions.

Every tool from this unit — matching bases, the inverse relationship between exponentials and logs, and the three log properties from 3.4 — exists to make one move possible: getting a variable out of an exponent or out from inside a logarithm. Which move you use depends on where the variable is stuck.

Exponential equations

Same base (or can be rewritten with one): if \(b^{m}\) = \(b^{n}\), then m = n. Rewrite both sides with a matching base, then set the exponents equal.
Different bases that don't simplify: take log or ln of both sides, then use the power rule to bring the exponent down as a multiplier, and divide to isolate the variable.

Logarithmic equations

Isolate the logarithm, then rewrite in exponential form using the definition from 3.3 (lo\(g_{b}\)(x) = y ⇔ \(b^{y}\) = x).
Multiple logs on one side: condense them into a single logarithm first (product/quotient rules from 3.4), then convert to exponential form.
Always check solutions against the domain of the original logarithms — a solution that makes any argument zero or negative is extraneous and must be discarded.

Solving 2ˣ = 10 graphically: the exponential meets y = 10 exactly where x = log₂10 ≈ 3.32

Worked Example 1 · Matching bases

ProblemSolve \(3^{2x-1}\) = 27.

1Rewrite 27 as a power of 3: 27 = 3³. Now both sides share base 3: \(3^{2x-1}\) = 3³.

2Set the exponents equal: 2x − 1 = 3, so 2x = 4, x = 2.

x = 2

Worked Example 2 · Different bases

ProblemSolve \(5^{x}\) = 40.

140 isn't a clean power of 5, so take ln of both sides: ln(\(5^{x}\)) = ln(40).

2Power rule brings the exponent down: x·ln(5) = ln(40).

3Divide: x = \(\frac{ln(40)}{ln(5)}\) ≈ \(\frac{3.689}{1.609}\) ≈ 2.292.

x ≈ 2.292

Worked Example 3 · Logarithmic equation with extraneous root

ProblemSolve log(x) + log(x − 3) = 1.

1Condense with the product rule: log(x(x−3)) = 1.

2Convert to exponential form (base 10): x(x−3) = 10¹ = 10, so x² − 3x − 10 = 0.

3Factor: (x−5)(x+2) = 0, so x = 5 or x = −2.

4Check both in the original equation: x = −2 makes log(−2) undefined, so it's extraneous. x = 5 gives log(5) + log(2), both defined.

x = 5 (x = −2 rejected)

Guided practice

1.Solve \(4^{x+1}\) = 64.
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64 = 4³, so x + 1 = 3. x = 2.
2.Solve \(2^{x}\) = 15, giving x to three decimal places.
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x = \(\frac{ln(15)}{ln(2)}\) ≈ \(\frac{2.708}{0.693}\) ≈ 3.907.
3.Solve lo\(g_{2}\)(x) = 5.
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Rewrite in exponential form: x = 2⁵. x = 32.
4.Solve ln(x) + ln(3) = ln(15).
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Condense: ln(3x) = ln(15). Since ln is one-to-one, 3x = 15, so x = 5.
5.Solve log(x) + log(x + 3) = 1, checking for extraneous solutions.
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Condense: log(x(x+3)) = 1, so x² + 3x = 10, x² + 3x − 10 = 0, (x+5)(x−2) = 0, giving x = −5 or x = 2. x = −5 makes log(−5) undefined, so it's rejected. x = 2.

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