Mathematics 9709/21 — October/November 2016

Cambridge AS Level · Pure Mathematics 2 · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Logarithmic and Exponential Functions · Trigonometry · Algebra · Differentiation · Integration · Numerical Solution of Equations

Q1Logarithmic and Exponential FunctionsAlgebraFree sample

(i)

It is given that $x$ satisfies the equation $3^{2x} = 5(3^x) + 14$ . Find the value of $3^x$ and, using logarithms, find the value of $x$ correct to 3 significant figures.

DifficultyMedium-Easy

Worked solution

Approach

Let $y = 3^x$ . Then $3^{2x} = (3^x)^2 = y^2$ , so the equation becomes a quadratic in $y$ . Solve it, take the positive root because $3^x > 0$ , then use logarithms to find $x$ .

Working

Let $y = 3^x$ . Since $3^{2x} = (3^x)^2$ :

y^2 = 5y + 14

Bring all terms to one side:

y^2 - 5y - 14 = 0

Factorise:

(y - 7)(y + 2) = 0

So:

y = 7 \quad \text{or} \quad y = -2

But $3^x > 0$ for all real $x$ , so:

3^x = 7

Taking logarithms:

x \ln 3 = \ln 7

x = \frac{\ln 7}{\ln 3}

Evaluating:

x \approx 1.7712\ldots

to 3 significant figures:

x = 1.77

Answer

Value of $3^x$ is $7$ ; value of $x$ is $1.77$ (3 s.f.).

Final answer

3^x = 7; x = 1.77 (3 s.f.)

Detailed explanation

Walkthrough

We first notice that $3^{2x}$ can be written as $(3^x)^2$ . Making the substitution $y = 3^x$ turns the equation into a quadratic $y^2 = 5y + 14$ , which is easier to solve than an exponential equation. Factorising gives $y = 7$ or $y = -2$ . Because an exponential function $3^x$ is always positive, only $y = 7$ is possible, so $3^x = 7$ . To get $x$ out of the exponent, take logarithms of both sides: this uses $\ln\left(3^x\right) = x\ln 3$ . Dividing by $\ln 3$ gives $x = \frac{\ln 7}{\ln 3}$ , and evaluating on a calculator gives $1.77$ to 3 significant figures.

Key Takeaways

An equation with $3^{2x}$ is a quadratic in disguise if we set $y = 3^x$ . Always reject negative values of $y$ because an exponential expression is positive. Taking logarithms is the standard way to solve an equation of the form $a^x = k$ .

Common Mistakes

Forgetting to reject $3^x = -2$ .
Solving $3^x = 7$ without using logarithms, or using the logarithm laws incorrectly.
Rounding to $1.8$ instead of $1.77$ ; the question asks for 3 significant figures.

Things to Be Careful About

Keep the substitution explicit so the quadratic factorisation is clear. When taking logarithms, apply $\ln\left(3^x\right) = x\ln 3$ , not $3\ln x$ . Use enough decimal places in the calculator value before rounding so the final 3 significant figures are accurate.

Techniques used

substitute y = 3^x to form a quadraticsolve the quadratic by factorisationtake logarithms of both sides to isolate the exponentround to the required significant figures

(ii)

Hence state the values of $x$ satisfying the equation $3^{2|x|} = 5(3^{|x|}) + 14$ .

DifficultyEasy

Worked solution

Approach

Put $u = |x|$ . The equation then has exactly the same form as the one in part (i). From part (i), the positive value of the exponential expression is $3^u = 7$ , so $|x| = \log_3 7$ . Since the modulus of $x$ is a positive constant, $x$ can be either plus or minus that constant.

Working

Let $u = |x|$ . The equation

3^{2|x|} = 5(3^{|x|}) + 14

has the same shape as the equation in part (i), so by the result of part (i):

3^{|x|} = 7

Thus

|x| = \log_3 7

From part (i),

\log_3 7 = 1.7712\ldots

x = 1.7712\ldots \quad \text{or} \quad x = -1.7712\ldots

To 3 significant figures:

x = 1.77 \quad \text{or} \quad x = -1.77

Answer

$x = \pm 1.77$ (3 s.f.).

Final answer

x = ±1.77 (3 s.f.)

Detailed explanation

Walkthrough

The new equation is identical in form to part (i), except that the variable $x$ has been replaced by $|x|$ . Therefore we can reuse the result without solving again: the positive value obtained in part (i) tells us that $3^{|x|} = 7$ . Taking logarithms gives $|x| = \log_3 7$ , approximately $1.77$ to 3 significant figures. The modulus equation $|x| = k$ has two solutions, $x = k$ and $x = -k$ , because both $k$ and $-k$ have modulus $k$ . Hence the two values are $x = \pm 1.77$ .

Key Takeaways

When an equation contains $|x|$ in the same position as $x$ in a previously solved equation, the same solution method applies with $u = |x|$ . Solving $|x| = k$ for $k > 0$ always gives two values.

Common Mistakes

Stating only $|x| = 1.77$ instead of giving the actual values of $x$ .
Forgetting the negative solution.
Using the negative root from part (i), since $3^{|x|}$ cannot equal $-2$ .

Things to Be Careful About

This is a follow-through mark, so use the positive answer from part (i). Keep the final values to 3 significant figures. Do not write the answer as only $|x| = 1.77$ ; state both signs for $x$ .

Techniques used

let u = |x| and compare with the solved equation in part (i)apply the modulus equation |u| = k to obtain the two signs

The rest of this paper

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