CAIE IGCSE Additional Math 4.3 Use Substitution for Related Equations Questions

Practise Cambridge IGCSE Additional Mathematics by transforming exponential and logarithmic equations, checking domains and reporting solutions.

Exam points

Solve exponential equations using substitutions, exponent laws and valid transformations.
Solve logarithmic equations using change of base, reciprocal relationships and domain restrictions.
Check transformed solutions in the original equation and report exact or appropriately rounded values.

[Maximum number: 6]

Solve the following equations.

$e^{3y-7}+\frac{4}{e^3}=\frac{5}{e^{3y-1}}$

$\begin{aligned} e^{3y-7+3y-1}+4e^{-3+3y-1}&=5 \quad\text{oe}\\ \text{or } e^{3y-4+3y-1}+4e^{3y-1}&=5e^3\\ \text{or } e^{3y-7}+\frac{4}{e^3}&=5e^{1-3y}\\ \text{or } e^{6y}+4e^{4+3y}&=5e^8 \end{aligned}$ $\begin{aligned} e^{6y-8}+4e^{3y-4}-5&=0\\ \text{or } e^{3y-4}+4-5e^{4-3y}&=0\\ \text{or } e^{-7}(e^{3y})^2+4e^{-3}e^{3y}-5e&=0\\ \text{or } (e^{3y-1})^2+4e^3e^{3y-1}-5e^6&=0 \end{aligned}$
B1 dep for arranging as 3-term quadratic.
$\begin{aligned} (e^{3y-4}-1)(e^{3y-4}+5)&=0\\ \text{or }(e^{-3}e^{3y}-e)(e^{-4}e^{3y}+5)&=0\\ \text{or }(e^{3y-1}+5e^3)(e^{3y-1}-e^3)&=0 \end{aligned}$
M1 dep for an attempt at solution.
$e^{3y-4}=1$ or $e^{3y-1}=e^3$ $3y-4=\ln 1$ oe soi or 3y-1=3 $y=\frac{4}{3}$

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CAIE IGCSE Additional Math 4.3 Use Substitution for Related Equations Questions question 1