Solving Log & Exponential Equations

Every log/exponential equation in 0606 is one of four patterns. Identify the pattern, run its routine, check the solutions, the checking line is a mark more often than students believe.

Pattern 1, condense to a single log

$\log_2(x + 3) + \log_2(x - 4) = 3$ $\log_2[(x + 3)(x - 4)] = 3$ (laws) $\to (x + 3)(x - 4) = 2^3 = 8$ $x^2 - x - 20 = 0 \to (x - 5)(x + 4) = 0 \to x = 5$ or $x = -4$ Check: $x = -4$ makes both arguments negative, rejected, with the reason written. $x = 5$

Condense (M) → definition (M) → solve (A) → reject (B). The rejection is non-negotiable: log equations manufacture invalid roots by design.

Pattern 2, log on both sides, same base

If $\log_a (f(x)) = \log_a (g(x))$, drop the logs: $f(x) = g(x)$, then solve and still check arguments are positive. Only valid when both sides are single logs of the same base; condense first if not.

Pattern 3, unknown in the exponent: take logs

$3^{2x+1} = 40$ $(2x + 1) \log 3 = \log 40 \to 2x + 1 = \dfrac{\log 40}{\log 3} \to$ $x = \dfrac{1}{2}\left(\dfrac{\log 40}{\log 3} - 1\right)$

Take logs of both sides, bring the exponent down with the power law, isolate. Any base of log works (lg or $\ln$); exact form is the Paper 1 expectation. If both sides are powers of a common base (e.g. $9^x = 27^{2-x} \to 3^{2x} = 3^{6-3x}$), equate exponents instead, faster and cleaner: $2x = 6 - 3x \to x = \dfrac{6}{5}$.

Pattern 4, disguised quadratic

$2^{2x} - 9(2^x) + 8 = 0$ Let $u = 2^x$ (the substitution line is the M mark): $u^2 - 9u + 8 = 0 \to u = 1, 8$ $2^x = 1 \to x = 0$; $2^x = 8 \to x = 3$. $x = 0$ or $3$

Spot it by the double-exponent structure: $a^{2x}$ is $(a^x)^2$. Negative $u$-values must be rejected in writing ($a^x > 0$ always). Finishing at $u = 1, 8$ without returning to $x$ is the classic dropped final mark.

Common mistakes

• Solutions never checked against positive-argument requirements
• Logs taken of one side only
• $\dfrac{\log 40}{\log 3}$ “simplified” to $\log\left(\dfrac{40}{3}\right)$, a fake law
• Common-base opportunities missed, leading to needless log arithmetic
• Disguised quadratics left in $u$

Full topic context: Logs & Exponentials notes · the e/ln versions: e^x and ln x.