Solving exponential equations using logarithms: Worked examples

The scenario matrix

Before solving, let's list every case class an exponential equation can belong to. Think of this as a checklist — each worked example below fills in one row.

Cell	Case class	What makes it different	Example below
A	Clean base, positive answer	$b^{x} = c$ with $c > b > 1$	Ex 1
B	Coefficient in front	$a \cdot b^{x} = c$ — isolate first	Ex 2
C	Negative exponent answer	$c < 1$ so the log is negative	Ex 3
D	Bracketed exponent	$b^{p x + q} = c$ — bracket the lump	Ex 4
E	Common base (no logs needed)	both sides are powers of one base	Ex 5
F	Unknown on both sides	logs on both, collect $x$ terms	Ex 6
G	Degenerate / no solution	$b^{x} =$ a non-positive number	Ex 7
H	Real-world growth/decay word problem	half-life / doubling time	Ex 8
I	Exam twist — quadratic in disguise	substitution $u = b^{x}$	Ex 9
J	Boundary case	$b^{x} = 1 \Rightarrow x = 0$	Ex 10

The single picture that governs all cases

Figure — Solving exponential equations using logarithms

Keep this picture beside you — every "Forecast" below is really asking which side of $x = 0$ will the answer land on?

Worked examples

Worked example Example 5 — Cell E: common base, no logs,

$4^{x + 1} = 3 2^{x - 1}$ Forecast: both $4$ and $32$ are powers of $2$ . If we rewrite everything in base $2$ , no decimals appear — expect a neat exact answer.

Step 1. Write in base $2$ : $4 = 2^{2}$ so $4^{x + 1} = 2^{2 (x + 1)} = 2^{2 x + 2}$ ; and $32 = 2^{5}$ so $3 2^{x - 1} = 2^{5 (x - 1)} = 2^{5 x - 5}$ . Why? Matching bases turns the problem into "equal bases ⇒ equal exponents", avoiding logs entirely.

Step 2. $2^{2 x + 2} = 2^{5 x - 5} \Rightarrow 2 x + 2 = 5 x - 5$ . Why can we equate exponents? Because $2^{()}$ is a one-to-one function (Inverse Functions) — different exponents give different values, so equal values force equal exponents.

Step 3. $2 + 5 = 5 x - 2 x \Rightarrow 7 = 3 x \Rightarrow x = \frac{7}{3} \approx 2.3333$ . Why? Ordinary linear algebra — gather $x$ -terms on one side, constants on the other, then divide by $3$ .

Verify: $4^{7/3 + 1} = 4^{10/3}$ and $3 2^{7/3 - 1} = 3 2^{4/3}$ . Both equal $2^{20/3} \approx 101.6$ ✓.

Worked example Example 6 — Cell F: unknown on both sides,

$3^{x} = 5^{x - 2}$ Forecast: bases differ ( $3$ vs $5$ ) and can't share a common integer base, so logs are unavoidable. We must gather all $x$ 's onto one side.

Step 1. $lo g (3^{x}) = lo g (5^{x - 2})$ . Why? Apply $lo g$ to both sides — since the bases differ we cannot match them, so logs are the only way to unstick both exponents at once.

Step 2. $x lo g 3 = (x - 2) lo g 5$ . Why bracket $(x - 2)$ ? The whole exponent on the right drops as a lump via the power law.

Step 3. Expand and collect $x$ : $x lo g 3 = x lo g 5 - 2 lo g 5$ , so $x lo g 3 - x lo g 5 = - 2 lo g 5$ , i.e. $x (lo g 3 - lo g 5) = - 2 lo g 5$ . Why? We factor $x$ out so we can divide it free — treat $lo g 3$ and $lo g 5$ as ordinary numbers.

Step 4. $x = \frac{- 2 lo g 5}{lo g 3 - lo g 5} = \frac{- 2 ( 0.6990 )}{0.4771 - 0.6990} = \frac{- 1.3979}{- 0.2218} \approx 6.302$ . Why? $(lo g 3 - lo g 5)$ is just a constant coefficient of $x$ — divide it off to isolate $x$ .

Verify: $3^{6.302} \approx 1006$ and $5^{6.302 - 2} = 5^{4.302} \approx 1006$ ✓.

Worked example Example 8 — Cell H: real-world decay, carbon-14 half-life

A sample of carbon-14 decays so that after $t$ years the fraction remaining is $N = (\frac{1}{2})^{t /5730}$ (its half-life is $5730$ years, see Exponential Growth and Decay). How long until only $30%$ remains?

Forecast: after $5730$ yr we're at $50%$ ; $30%$ is less than $50%$ , so it takes longer than one half-life. Guess $t$ somewhat above $5730$ .

Step 1. Set up: $(\frac{1}{2})^{t /5730} = 0.30$ . Why? "Fraction remaining $= 30%$ " means the decay expression equals $0.30$ — translate the words into an equation.

Step 2. $lo g ((\frac{1}{2})^{t /5730}) = lo g 0.30$ . Why? The exponential term is already alone, so apply $lo g$ to both sides to expose the exponent.

Step 3. $\frac{t}{5730} lo g (0.5) = lo g (0.30)$ . Why bracket $t /5730$ ? It is the whole exponent — it drops as one piece via the power law.

Step 4. $\frac{t}{5730} = \frac{lo g 0.30}{lo g 0.5} = \frac{- 0.5229}{- 0.3010} \approx 1.7370$ . Why? $lo g 0.5$ is a constant multiplying the exponent — divide it off. (Two negatives make a positive here.)

Step 5. $t \approx 1.7370 \times 5730 \approx 9953$ years. Why? Multiply through by $5730$ to undo the division and free $t$ .

Verify (units + sanity): $t$ is in years ✓. It exceeds one half-life ( $5730$ yr) because we passed below $50%$ ✓. Plug back: $(0.5)^{9953/5730} = (0.5)^{1.737} \approx 0.300 = 30%$ ✓.

Worked example Example 9 — Cell I: exam twist, quadratic in disguise,

$2^{2 x} - 5 \cdot 2^{x} + 4 = 0$ Forecast: the two exponential terms are $2^{2 x}$ and $2^{x}$ . Notice $2^{2 x} = (2^{x})^{2}$ . So if we call $2^{x}$ a single letter, this becomes an ordinary quadratic — expect two answers.

Step 1. Substitute $u = 2^{x}$ . Then $2^{2 x} = (2^{x})^{2} = u^{2}$ , giving $u^{2} - 5 u + 4 = 0$ . Why substitute? It converts a scary exponential into a familiar quadratic we can factor.

Step 2. Factor: $(u - 1) (u - 4) = 0 \Rightarrow u = 1$ or $u = 4$ . Why? A product equals zero only when one factor is zero — the standard route to a quadratic's roots.

Step 3. Undo the substitution — back to exponentials:

$u = 1$ : $2^{x} = 1 \Rightarrow x = 0$ (since $2^{0} = 1$ ).
$u = 4$ : $2^{x} = 4 \Rightarrow x = 2$ (since $2^{2} = 4$ ).

Why check both? A valid $u$ must be positive (as $2^{x} > 0$ always). Here both $1$ and $4$ are positive, so both survive. Had a root been $u \leq 0$ , we would reject it — a common exam trap.

Verify: at $x = 0$ : $2^{0} - 5 \cdot 2^{0} + 4 = 1 - 5 + 4 = 0$ ✓. At $x = 2$ : $2^{4} - 5 \cdot 2^{2} + 4 = 16 - 20 + 4 = 0$ ✓.

Recall Quick self-test across the whole matrix

Which cell does $9^{x} = 1$ hit, and what is $x$ ? ::: Boundary case J ( $c = 1$ ): $x = 0$ , because any base to the power $0$ is $1$ . In $4 \cdot 2^{x} = 100$ , why divide by 4 before logging? ::: The power law only frees the exponent when the exponential is isolated; the coefficient $a = 4$ is a separate multiplier (Cell B). Why does $2^{x} = - 8$ have no solution? ::: $b^{x}$ is always positive; the graph never dips to a negative value, so the line $y = - 8$ never meets it (Cell G). In $2^{2 x} - 5 \cdot 2^{x} + 4 = 0$ , what substitution unlocks it? ::: Let $u = 2^{x}$ , giving a quadratic $u^{2} - 5 u + 4 = 0$ (Cell I). When should you match bases instead of using logs? ::: When both sides are powers of a common base — it gives an exact answer with no decimals (Cell E).

Connections

Solving exponential equations using logarithms — the parent recipe these examples drill.
Laws of Logarithms — the power law behind every "drop the exponent" step.
Change of Base Formula — every answer $x = \frac{l o g c}{l o g b}$ is $lo g_{b} c$ .
Inverse Functions — why equal bases force equal exponents (Ex 5) and why $b^{x} > 0$ (Ex 7).
The Number e and Natural Logarithms — swap $lo g$ for $ln$ ; answers are identical.
Exponential Growth and Decay — home of the half-life problem (Ex 8).
Solving Equations Graphically — the master picture: line meets curve.

Worked examples — Solving exponential equations using logarithms