9.4 Solving Exponential and Logarithmic Equations

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.

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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

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.

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