(i) Use logarithms to solve the equation $2^x = 20^5$, giving the answer correct to 3 significant figures. (ii) Hence determine the number of integers $n$ satisfying $$ 20^{-5} < 2^n < 20^5 $$

Mathematics 9709/22 — May/June 2015

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

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

Q1Logarithmic and Exponential FunctionsFree sample

(i)

Use logarithms to solve the equation $2^x = 20^5$ , giving the answer correct to 3 significant figures.

DifficultyMedium-Easy

Worked solution

Approach

Take logarithms of both sides so that the unknown index $x$ can be brought down using the power law, then solve for $x$ and round to 3 significant figures.

Working

Take logarithms of both sides:

\log\left(2^x\right) = \log\left(20^5\right)

Use the power law $\log(a^b) = b\log a$ :

x\log 2 = 5\log 20

Divide by $\log 2$ :

x = \frac{5\log 20}{\log 2}

Evaluating with a calculator:

x = 21.6096404744\ldots

Rounded to 3 significant figures:

x = 21.6

Answer

x = 21.6

Final answer

21.6

Detailed explanation

Walkthrough

The equation has the unknown in the exponent, so ordinary algebraic manipulation cannot isolate it directly. Taking logarithms of both sides is the standard technique because the logarithm of a power allows the exponent to be brought down as a multiplier. We apply the power law $\log(a^b) = b\log a$ to obtain $x\log 2 = 5\log 20$ . Dividing both sides by $\log 2$ gives $x = \frac{5\log 20}{\log 2}$ . A calculator gives $21.6096404744\ldots$ , and rounding to 3 significant figures gives $21.6$ .

Key Takeaways

This question tests the fundamental logarithm–index relationship: if $a^b = c$ , then $b\log a = \log c$ . It also tests the power law of logarithms and the ability to round to a specified number of significant figures.

Common Mistakes

Forgetting to apply the power law and leaving $\log(2^x)$ without bringing $x$ down.
Using different bases in the two logarithms; any base is acceptable, but both logarithms must use the same base.
Rounding too early and losing accuracy; keep the full calculator value before rounding.
Giving the answer as $21.609$ instead of $21.6$ when 3 significant figures are requested.

Things to Be Careful About

Make sure both sides are logged, not just the left side. Also remember that in $\log(20^5)$ , the exponent $5$ multiplies $\log 20$ , so the numerator is $5\log 20$ , not $\log 100$ or similar. Finally, 3 significant figures means three digits total, so $21.6$ has three significant figures.

Techniques used

take logarithms of both sidesapply the power law of logarithmsdivide to isolate the unknown indexround to the required number of significant figures

(ii)

Hence determine the number of integers $n$ satisfying

20^{-5} < 2^n < 20^5

DifficultyMedium-Easy

Worked solution

Approach

Use the result from part (i) to rewrite the bounds $20^5$ and $20^{-5}$ in terms of powers of $2$ . Since $2^n$ is an increasing function, the inequality becomes bounds on $n$ . Then count the integers in that interval.

Working

From part (i), $2^x = 20^5$ with $x = 21.609\ldots$ , so:

20^5 \approx 2^{21.6}

and therefore

20^{-5} = \frac{1}{20^5} \approx 2^{-21.6}

The inequality is:

2^{-21.6} < 2^n < 2^{21.6}

Since $2^n$ is strictly increasing, this is equivalent to:

-21.6 < n < 21.6

The integers $n$ satisfying this are:

-21, -20, \ldots, 21

The number of integers from $-21$ to $21$ inclusive is:

21 - (-21) + 1 = 43

Answer

43

Final answer

Detailed explanation

Walkthrough

Part (i) found the value of $x$ such that $2^x = 20^5$ , namely $x = 21.609\ldots$ . This lets us rewrite the upper bound $20^5$ as approximately $2^{21.6}$ . For the lower bound, $20^{-5}$ is the reciprocal of $20^5$ , so it equals approximately $2^{-21.6}$ . The inequality becomes $2^{-21.6} < 2^n < 2^{21.6}$ . Since the exponential function $2^n$ is strictly increasing, we can compare exponents directly: $-21.6 < n < 21.6$ . The integers in this open interval are from $-21$ to $21$ inclusive. To count them, use $21 - (-21) + 1 = 43$ .

Key Takeaways

This part connects an equation result to an inequality. It also reinforces that the exponential function is increasing, so the direction of the inequality is preserved when comparing exponents. Counting integers in an interval requires checking the endpoints carefully.

Common Mistakes

Forgetting that the lower bound is negative and writing only $n < 21.6$ .
Counting from $0$ to $21$ and doubling incorrectly, or forgetting to include $0$ .
Using inclusive endpoints and counting $22$ or $44$ instead of $43$ .
Confusing $20^{-5}$ with $-20^5$ ; the negative exponent means reciprocal, not a negative number.

Things to Be Careful About

The inequalities are strict: $2^n < 20^5$ and $2^n > 20^{-5}$ . Since $21.6$ is not an integer, the strictness does not exclude the endpoints $-21$ and $21$ , so both are included. The count from $-21$ to $21$ inclusive is $43$ , because there are $21 - (-21) + 1$ integers.

Techniques used

rewrite bounds using the result from part (i)apply the monotonicity of the exponential functioncount integers in an interval

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