How Do You Solve an Equation Like e^(2x) - 5e^x + 4 = 0?

Use the substitution $u=e^x$ to convert the exponential equation into a quadratic. Solving it gives the two real solutions $\boxed{x=0}$ and $\boxed{x=\ln 4}$.

The Analytical Method

This is a hidden quadratic because

$$e^{2x}=\left(e^x\right)^2.$$

The exponents are in the ratio $2:1$, so the larger exponential term is the square of the smaller one. This pattern signals that one substitution can produce a standard quadratic equation.

Why the Substitution Works

The substitution is valid because the exponential function maps every real $x$ to a positive value. It also turns powers of the same expression into ordinary powers of $u$. More generally, an equation of the form $ae^{2x}+be^x+c=0$ can be rewritten as , where . Solve the quadratic first, reject every non-positive value of , and then solve each remaining equation using . This approach is usually faster and clearer than attempting to apply logarithms at the start. It also preserves exact answers.

Start with

$$e^{2x}-5e^x+4=0.$$

Let $u=e^x$. Then $e^{2x}=u^2$, giving

$$u^2-5u+4=0.$$

Factorize the quadratic:

$$(u-1)(u-4)=0.$$

Therefore, $u=1$ or $u=4$. Substituting back:

$$e^x=1 \Rightarrow x=\ln 1=0,$$ $$e^x=4 \Rightarrow x=\ln 4.$$

Hence,

$$\boxed{x=0 \text{ or } x=\ln 4}.$$

The domain restriction $u=e^x>0$ must be checked. Both roots are positive, so both are valid. A negative root would be rejected.

Common Misconception

Logarithms do not distribute over addition: $\ln(a+b)\neq \ln a+\ln b$. Use logarithms only after isolating a positive exponential expression.

IB Exam Technique

This is IB Mathematics: Analysis and Approaches, SL 2.10: solving equations analytically. It suits Paper 1 because the solutions are exact. Show the substitution, factorization, domain check, and back-substitution; an unsupported answer may not earn full marks.