Exp and log equations: e^x = k and ln(x) = k in 30 seconds

Exponential equation: $e^x = k$

• If $k > 0$: unique solution $x = \ln(k)$.
• If $k \leq 0$: no solution (since $e^x > 0$ always).

Example: $e^x = 5$ ⇒ $x = \ln(5) \approx 1.609$.

Logarithmic equation: $\ln(x) = k$

Domain: $x > 0$.

Always a unique solution: $x = e^k$.

Example: $\ln(x) = 2$ ⇒ $x = e^2 \approx 7.389$.

More complex cases

$e^{ax + b} = k$ with $k > 0$: $a x + b = \ln(k) \Rightarrow x = \dfrac{\ln(k) - b}{a}$

$\ln(a x + b) = k$ with $a x + b > 0$: $a x + b = e^k \Rightarrow x = \dfrac{e^k - b}{a}$

⚠️ Then verify the condition $a x + b > 0$.

Inequalities

$e^x > k$ (exp is strictly increasing):

• If $k \leq 0$: $S = \mathbb{R}$.
• If $k > 0$: $x > \ln(k)$.

$\ln(x) > k$ (ln is strictly increasing):

• Always $x > e^k$ (and $x > 0$ automatic).

Solving $e^{f(x)} = e^{g(x)}$

Since $\exp$ is injective: $f(x) = g(x)$.

Same for $\ln$: $\ln(f(x)) = \ln(g(x)) \iff f(x) = g(x)$ with $f(x), g(x) > 0$.

Common mistakes

1. Solving $e^x = -3$ without saying “no solution”.
2. Forgetting the condition $x > 0$ for $\ln$.
3. Writing $\ln(a) + \ln(b) = \ln(a + b)$ (wrong: it’s $\ln(ab)$).
4. Writing $e^{a+b} = e^a + e^b$ (wrong: it’s $e^a \cdot e^b$).

?À toi de jouer

Solution of e^(2x − 1) = 5?

A.(ln(5) + 1) / 2B.ln(5) − 1C.(5 + 1) / 2D.e^5 / 2

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