Study guide
9.4 Solving Exponential and Logarithmic Equations
ACCUPLACER Advanced Algebra and Functions Exam
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1 Introduction & Exam Overview
1.1 Current Exam Facts 1.2 Scoring, Skills Insight Bands, and Placement 1.3 How to Study for AAF
2 Linear Equations
2.1 Writing Linear Equations in One and Two Variables 2.2 Solving Linear Equations 2.3 Linear Inequalities 2.4 Systems of Two Linear Equations
3 Linear Applications and Graphs
3.1 Slope, Intercepts, and Graphing Lines 3.2 Parallel and Perpendicular Lines 3.3 Modeling with Linear Functions 3.4 Graphing Linear Inequalities and Systems
4 Factoring
4.1 GCF, Grouping, and Special Products 4.2 Factoring Quadratic Trinomials 4.3 Factoring Cubics and Higher Polynomials
5 Quadratics
5.1 Quadratic Forms, Graphs, Vertex, and Intercepts 5.2 Solving Quadratics by Factoring 5.3 Quadratic Formula and Completing the Square 5.4 Quadratic Inequalities and Quadratic-Linear Systems
6 Functions
6.1 Function Notation, Evaluation, Domain, and Range 6.2 Graphs of Functions, Vertical Line Test, Maxima and Minima 6.3 Function Operations, Composition, and Inverses 6.4 Transformations and Interpreting Functions in Context
7 Radical and Rational Equations
7.1 Simplifying Radical Expressions 7.2 Solving Radical Equations and Extraneous Roots 7.3 Rational Expressions and Equations 7.4 Graphs, Domain, and Range of Radical and Rational Functions
8 Polynomial Equations
8.1 Polynomial Operations and Structure 8.2 Solving Polynomial Equations 8.3 Graphing Polynomials, Roots, and End Behavior
9 Exponential and Logarithmic Equations
9.1 Exponential Functions, Growth, and Decay 9.2 Graphs of Exponential Functions 9.3 Logarithms and Logarithmic Properties 9.4 Solving Exponential and Logarithmic Equations
10 Geometry Concepts for Algebra 1
10.1 Area, Perimeter, and Prism Volume 10.2 Distance Formula and Pythagorean Theorem 10.3 Dilations, Rotations, Translations, and Reflections
11 Geometry Concepts for Algebra 2
11.1 Volume of Cylinders, Cones, and Spheres 11.2 Intersecting Lines and Angle Relationships 11.3 Triangle Similarity and Congruence 11.4 Circles in the Coordinate Plane
12 Trigonometry
12.1 Right-Triangle Trigonometry and Special Triangles 12.2 Unit Circle, Radians, and Arc Length 12.3 Trigonometric Graphs and Identities 12.4 Solving Trig Equations; Law of Sines and Cosines
9.4 Solving Exponential and Logarithmic Equations
Key Takeaways Explain This Topic AI
Same-base rule: if b^u = b^v with b > 0, b ≠ 1, then u = v. Example: 9^x = 27 becomes (3^2)^x = 3^3, so 2x = 3 and x = 3/2.
When bases will not match, convert: 3^{4x} = 11 implies 4x = log_3 11, so x = (log_3 11)/4, equivalently ln 11 / (4 ln 3).
Log equations: rewrite log_b(argument) = c as argument = b^c, then reject any candidate that makes an argument ≤ 0.
Worked domain check: log_2 (x − 3) = 4 gives x − 3 = 16, so x = 19, and 19 − 3 = 16 > 0, so x = 19 holds.
Table 11 solving is 5–15% of AAF (1–3 CAT items) paired with creating, graphing, and interpreting; leave exact log forms unless the options are decimals.
Last updated: August 2026
9.4 Solving Exponential and Logarithmic Equations
Table 11’s solving skill for this 5–15% content area (typically 1–3 CAT items) is algebraic: find the input that makes an exponential or a log statement true. The definition from Logarithms and Logarithmic Properties is the engine — log_b a = c iff b^c = a — together with same-base exponent laws. This section is the workout: same-base exponentials, converting when bases will not match, exponentiating a log equation, and checking domain so extraneous roots do not survive.
Creating y = a b^t from a story was Exponential Functions, Growth, and Decay . Work FREE mixed AAF items at /practice/accuplacer-advanced-algebra . Official skill names are on College Board’s What’s on the Tests page.
Same-base exponential equations
If b > 0 , b ≠ 1 , and b^u = b^v , then u = v . The exponential function is one-to-one, so equal outputs mean equal exponents. The method is: rewrite both sides as powers of the same base, then set the exponents equal.
Worked: 9^x = 27
Both 9 and 27 are powers of 3:
- 9^x = (3^2)^x = 3^{2x}
- 27 = 3^3
So 3^{2x} = 3^3 , hence 2x = 3 , and x = 3/2 .
Check: 9^{3/2} = (9^{1/2})^3 = 3^3 = 27 . The same check as (9^3)^{1/2} = 729^{1/2} = 27 . Wrong answers x = 3 (copied the 27’s exponent without rewriting 9), x = 2/3 (flipped 2x = 3), and x = 6 (multiplied 2 · 3 instead of dividing) are all on the table.
Worked: 5^{x + 1} = 125
125 = 5^3 , so 5^{x + 1} = 5^3 , hence x + 1 = 3 , and x = 2 . Check: 5^{3} = 125 . If the right side had been 25, you would get x + 1 = 2 , x = 1 .
Worked: 4^{2x − 1} = 8^{x + 1}
Rewrite 4 and 8 as powers of 2: 4 = 2^2 and 8 = 2^3 .
- (2^2)^{2x − 1} = (2^3)^{x + 1}
- 2^{4x − 2} = 2^{3x + 3}
- 4x − 2 = 3x + 3
- x = 5
Check one side: 4^{9} versus 8^{6} is tedious by hand, but 4^{2·5 − 1} = 4^9 and 8^{5+1} = 8^6 , and in base 2 both exponents are 18 . The algebra is the item; AAF often stops at x = 5 .
If a coefficient sits in front, divide first when it helps: 2 · 3^x = 54 ⇒ 3^x = 27 ⇒ x = 3 . Do not take a log of the 2 and the 3^x separately without using the product rule; isolating the exponential is cleaner.
Bases that will not match: convert to a logarithm
College Board’s identity in this family is the move “if 2^{5x} = 7 , then x = (log_2 7)/5 .” Learn that move on original numbers so you are not parroting the sample. Use 3^{4x} = 11 .
3^{4x} = 11
Take log_3 of both sides (or convert by the definition):
- 4x = log_3 11
- x = (log_3 11)/4
Change of base if the options use ln or common log:
x = ln 11 / (4 ln 3) = log 11 / (4 log 3)
What is not correct:
x = 11/4 treats the 3 as if it disappeared.
x = log_3 (11/4) applies the log to a rewritten quotient that was never there.
x = 4 log_3 11 multiplies by 4 instead of dividing.
x = log_11 3 / 4 inverts the ratio.
The pattern: b^{kx} = c ⇒ kx = log_b c ⇒ x = (log_b c)/k . That is the solving skill paired with the definition, not a new formula to memorize in isolation.
Another original: 7^{x − 2} = 20 . Then x − 2 = log_7 20 , so x = 2 + log_7 20 . Do not drop the −2 .
If both sides already have the unknown in an exponent and the bases differ, take a log of both sides and use the power rule: 5^x = 3^{x+1} ⇒ x ln 5 = (x+1) ln 3 , then collect x terms. That is the same inverse idea, written with ln for convenience.
Logarithmic equations: exponentiate, then check domain
To solve log_b (expression) = c , rewrite as expression = b^c , then solve the resulting algebra (often linear or quadratic). Afterward, reject any candidate that makes an argument ≤ 0 .
Worked: log_2 (x − 3) = 4
Convert: x − 3 = 2^4 = 16 , so x = 19 .
Domain: need x − 3 > 0 , i.e. x > 3 . The candidate 19 satisfies 19 > 3 , and log_2 (16) = 4 checks. x = 19 is the solution. Wrong answers include x = 16 (forgot to add 3), x = 7 (used 2^3 or added 4 instead of 16), and x = 1 (solved x − 3 = −2 from a sign error).
Worked: log_3 x + log_3 (x − 2) = 1
Product rule: log_3 [x(x − 2)] = 1 .
Convert: x(x − 2) = 3^1 = 3 .
- x^2 − 2x − 3 = 0
- (x − 3)(x + 1) = 0
Candidates x = 3 and x = −1 .
Domain: need x > 0 and x − 2 > 0 , so x > 2 . (Each original argument must be positive, not only the product.) Reject x = −1 . Keep x = 3 . Check: log_3 3 + log_3 1 = 1 + 0 = 1 .
The extra root x = −1 is extraneous . It can satisfy the quadratic after you have multiplied arguments, but it was never in the domain of the original logs. This is the same extraneous-root discipline as radical equations in Solving Radical Equations and Extraneous Roots : algebra after a one-way rewrite can add junk; the original equation is the judge.
Worked: log_5 (2x + 1) = log_5 (x + 4)
If log_b u = log_b v with equal bases, then u = v provided both arguments stay positive. So 2x + 1 = x + 4 , x = 3 . Check: 2(3)+1 = 7 and 3+4 = 7 , both positive. If a candidate had made 2x + 1 ≤ 0 or x + 4 ≤ 0 , drop it even though the linear step produced it.
Mixed strategy and AAF traps
- Equation type First move Watch for
- b^u = b^v Set u = v Rewrite bases (9 and 27 → 3)
- b^{kx} = c x = (log_b c)/k Divide by k ; do not write log_b (c/k)
- log_b (u) = c u = b^c Then u > 0
- log u + log v = c log(uv) = c Domain of each factor
- log u = log v u = v Both u > 0 and v > 0
Traps that recycle across items:
Taking log(A + B) as log A + log B .
Forgetting domain and keeping a negative argument.
Treating 3^{4x} = 11 as 4x = 11 (dropping the base).
Writing (log_3 11)/4 as log_3 11 − 4 .
Solving log_2 (x − 3) = 4 as x − 3 = 4 (forgetting to raise 2 to the 4th).
Exact forms are preferred. x = (log_3 11)/4 is a finished AAF answer when that expression is in the options. Decimal approximations without a calculator icon waste time and may not match. Handheld calculators are not allowed except with an approved accommodation (College Board calculator policy).
Interpreting a solution in context closes Table 11’s four-skill loop. If a CD model 2500(1.06)^t = 4000 is solved as t = log(4000/2500) / log(1.06) , that t is years until the balance hits $4,000 — an interpreting step after solving , built on a function you created in 9.1. Graphically, the same t is the x-coordinate where y = 2500(1.06)^x meets the line y = 4000 .
When you can move among creating y = a b^t , graphing y = a^{x − h} + k , evaluating log_2 8 = 3 , and solving 3^{4x} = 11 as x = (log_3 11)/4 with a domain check on every log equation, this 5–15% slice of AAF is in working order. Keep rehearsing with FREE items at /practice/accuplacer-advanced-algebra .