Mathematical Methods — Question 16

QCAA 2023, Paper 1 · 5 marks

Q16 · 2023 · Technology-freeComplex familiar5 marks

QUESTION 16 (5 marks)

Solve for

x

in the equation

4+7e^{-2x}=3e^{2x}.

WORKED SOLUTION

5 marks

ANSWER

x=\dfrac12\ln\left(\dfrac73\right)

Worked solution

4+7e^{-2x}=3e^{2x}.

Multiplying by

e^{2x}

gives

4e^{2x}+7=3e^{4x},

3e^{4x}-4e^{2x}-7=0.

Factorising,

(3e^{2x}-7)(e^{2x}+1)=0.

The equation

e^{2x}=-1

is not possible, so

e^{2x}=\frac73.

Thus

\ln\left(\frac73\right)=2x\quad\Rightarrow\quad x=\frac12\ln\left(\frac73\right).

correctly removes the negative index in the equation

[1 mark]

rearranges equation to equate to zero

[1 mark]

factorises the equation

[1 mark]

rejects the non-feasible solution

[1 mark]

determines a feasible solution for x

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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