Exponential equation by quadratic substitution

Practise exponential equation by quadratic substitution with an original VectorPaths question, worked explanation and related VCE Methods practice.

4 practice marks

Original practice question

Solve $e^{2x}-5e^x+6=0$. (a) Use a substitution to obtain a quadratic. [2 marks] (b) Find all real x exactly. [2 marks]

Worked explanation

• Part (a)

  Set $u=e^x$. Then $e^{2x}=u^2$, so the equation becomes $u^2-5u+6=(u-2)(u-3)=0$.

• Part (b)

  Both $u=2$ and $u=3$ are positive. Taking natural logarithms gives the two real solutions.

(a) $u=e^x>0$, $u^2-5u+6=0$.; (b) $x=\ln2,\ln3$.

Original VectorPaths practice; not an official VCAA question or marking rubric.