Exponents and logarithms: exam review Exam review One question at a time

Use the calculator where it helps. [7]

1. (a) Solve \(10^x = 250\). Give x to 3 significant figures. [2]

   • M1 for x = log10 250
   • A1 2.40

2. (b) Solve \(\log_{10}(2x − 3) = 2\). Give the exact value of x. [2]

   • M1 for 2x − 3 = 10^2 = 100
   • A1 x = \frac{103}{2} (51.5)

3. (c) Solve \(7 − 3\ln x = 2\). Give the exact value of x, and x to 3 significant figures. [3]

   • M1 for isolating the log: ln x = \frac{5}{3}
   • M1 for x = e^(\frac{5}{3})
   • A1 5.29

Do not use a calculator for this question. [5]

1. (a) Write \(\dfrac{(3x^4)^2}{6x^{-3}}\) in the form \(ax^p\). Write down the values of a and p. [3]

   • M1 for (3x^4)^2 = 9x^8
   • M1 for dividing by x^(−3), so the indices are added: 8 + 3
   • A1 a = 3/2, p = 11

2. (b) Find the exact value of \(\left(\dfrac{3}{4}\right)^{-2}\). [2]

   • M1 for turning the fraction over: (4/3)^2
   • A1 16/9

Do not use a calculator for this question. [7]

1. (a) Find the exact value of \(32^{-\frac{3}{5}}\). [2]

   • M1 for 32^(1/5) = 2
   • A1 1/8

2. (b) Given that \(y=\dfrac{6}{x^2\sqrt{x}}\), write \(y^2\) in the form \(ax^k\). Write down the values of a and k. [3]

   • M1 for writing √x as x^(1/2), so y = 6x^(−2.5)
   • M1 for squaring, multiplying the index by 2
   • A1 a = 36, k = −5

3. (c) Write \(\dfrac{x^2 + 5\sqrt{x}}{x}\) in the form \(ax^p + bx^q\). [2]

   • M1 for dividing each term by x
   • A1 x + 5x^(−1/2)

Solve each equation. Do not use a calculator. [5]

1. (a) Solve \(x^{\frac{2}{3}} = 49\). [2]

   • M1 for raising both sides to the power 3/2: x = 49^(3/2)
   • A1 x = 343

2. (b) Solve \(\left(\sqrt{2}\right)^x = 16^{x-3}\). Give the exact value of x. [3]

   • M1 for writing both sides as powers of 2: 2^(x/2) = 2^(4(x − 3))
   • M1 for equating the indices: x/2 = 4x − 12
   • A1 x = 24/7

Give exact answers. Do not use a calculator. [5]

1. (a) Let \(p=\log_{10}a\) and \(q=\log_{10}b\). Write \(\log_{10}\left(\dfrac{10a^{3}}{\sqrt{b}}\right)\) in terms of p and q. [3]

   • M1 for the product or quotient law: log of a product is a sum, log of a quotient is a difference
   • M1 for the power law, including the root as a power 1/2
   • A1 1 + 3p − \frac{1}{2}q

2. (b) Write \(2 \ln 3 + \ln 20 − \ln 6\) as a single logarithm in the form \(\ln k\). [2]

   • M1 for the power law on the coefficient, then the product and quotient laws
   • A1 ln 30

Do not use a calculator for this question. [5]

1. (a) Find the exact value of \(\left(\frac{2}{3}\right)^{-4}\). [2]

   • M1 for flipping the fraction: \(\left(\frac{3}{2}\right)^{4}\)
   • A1 \(\frac{81}{16}\)

2. (b) Write \(\dfrac{(4x^2)^3}{8x^{-2}}\) in the form \(ax^p\). Write down the values of a and p. [3]

   • M1 for \((4x^2)^3=64x^{6}\)
   • M1 for dividing the numbers and subtracting the powers
   • A1 \(a = 8\), p = 8

Solve each equation. [5]

1. (a) Solve \(\ln(x+3)=2\). Give x to 3 significant figures. [2]

   • M1 for \(x+3=e^2\)
   • A1 \(x = e^2-3 = 4.39\)

2. (b) Solve \(2^{x+1}=8^{x-3}\). Give the exact value of x. [3]

   • M1 for writing \(8^{x-3}=2^{3(x-3)}\)
   • M1 for \(x+1=3x-9\)
   • A1 x = 5

A hard drive can store \(3.2\times10^{12}\) bytes. A photo file uses \(2.5\times10^{6}\) bytes. [5]

1. (a) Find how many photos the drive can store. Give your answer in the form \(a\times10^{k}\), where \(1\le a<10\) and \(k\in\mathbb{Z}\). [2]

   • M1 for \(\dfrac{3.2\times10^{12}}{2.5\times10^{6}}\) (\(1.28\times10^{6}\), the a values divided and the indices subtracted)
   • A1 \(1.28\times10^{6}\)

2. (b) A second drive can store \(8.5\times10^{11}\) bytes. Find the total number of bytes the two drives can store. Give your answer in the form \(a\times10^{k}\). [2]

   • M1 for writing both numbers with the same power of 10, for example \(3.2\times10^{12}+0.85\times10^{12}\)
   • A1 \(4.05\times10^{12}\)

3. (c) One gigabyte is \(10^{9}\) bytes. Write down the capacity of the first drive in gigabytes. [1]

   • A1 3200 (\(3.2\times10^{12}\div10^{9}=3.2\times10^{3}\))

The energy E, in joules, released by an earthquake of magnitude M satisfies \(\log_{10}E=5.2+1.5M\). [5]

1. (a) Find the energy released by an earthquake of magnitude 6. Give your answer in the form \(a\times10^{k}\), where \(1\le a<10\) and \(k\in\mathbb{Z}\), with a correct to 3 significant figures. [2]

   • M1 for \(\log_{10}E=5.2+1.5\times6=14.2\), so \(E=10^{14.2}\)
   • A1 \(1.58\times10^{14}\)

2. (b) An earthquake of magnitude 8 and an earthquake of magnitude 5 occur. Find how many times more energy the magnitude 8 earthquake releases. Give your answer to 3 significant figures. [3]

   • M1 for \(\log_{10}E_8-\log_{10}E_5=1.5\times(8-5)=4.5\) (the 5.2 cancels)
   • M1 for \(\log_{10}\dfrac{E_8}{E_5}=4.5\), so \(\dfrac{E_8}{E_5}=10^{4.5}\)
   • A1 3.16e+04

Solve each equation. [13]

1. (a) Solve \(\log_{10}(x+4) + \log_{10}5 = 3\). Give an exact answer. [3]

   • M1 for combining: log10 5(x + 4) = 3
   • M1 for 5(x + 4) = 10^3 = 1000
   • A1 x = 196

2. (b) Solve \(\ln(2x+1) - \ln(x-2) = 1\). Give x to 3 significant figures. [4]

   • M1 for combining: ln ((2x + 1)/(x − 2)) = 1
   • M1 for (2x + 1)/(x − 2) = e^1
   • M1 for collecting the terms in x and solving
   • A1 8.96

3. (c) Solve \(4^{x+1} = 7^{x-2}\). Give x to 3 significant figures. [4]

   • M1 for taking ln of both sides: (x + 1) ln 4 = (x − 2) ln 7
   • M1 for expanding and collecting the terms in x
   • M1 for x = 1 ln 4 + 2 ln 7 over ln 7 − ln 4
   • A1 9.43

4. (d) Solve \(\log_{x}81 = \frac{4}{3}\). [2]

   • M1 for x^(4/3) = 81
   • A1 x = 27

The number of users N (thousands) of an app, t months after it is launched, is modelled by \(N = 45\times1.12^{t}\). [8]

1. (a) Write down the value of N when t = 0, and find the value of N when t = 6. [2]

   • A1 N = 45
   • A1 88.8 (45 × 1.12^6)

2. (b) Use your calculator to find when N reaches 200. Give t to 3 significant figures. [2]

   • M1 for solving 45 × 1.12^t = 200
   • A1 13.2

3. (c) A second model is \(M = 120\times1.04^{t}\). Find the time when M = N. Give t to 3 significant figures. [2]

   • M1 for solving 45 × 1.12^t = 120 × 1.04^t
   • A1 13.2

4. (d) Find the time for N to multiply by 10. Give t to 3 significant figures. [2]

   • M1 for solving 1.12^t = 10
   • A1 20.3

A biologist studies how the metabolic rate y (watts) of an animal depends on its body mass x (kg). The masses in the study range from 0.03 kg to 5000 kg. [13]

1. (a) Write down the logarithm to base 10 of each of the values 0.03, 12, 5000. Give each to 3 significant figures. [2]

   • M1 for using log10 of the values
   • A1 −1.52, 1.08, 3.70

2. (b) Give one reason why a logarithmic scale is useful for data like this. [1]

   • R1 the values cover several powers of 10, so the logs are of similar size and can be shown on one graph

3. (c) For the data in the table, the correlation coefficient for \(\log_{10}y\) against x is r = 0.7546, and for \(\log_{10}y\) against \(\log_{10}x\) it is r = 0.9996. Decide whether an exponential model or a power model fits the data better. Give a reason. [2]

   • M1 for comparing the two values of r (the one closer to 1 is nearer a straight line)
   • A1 a power model, \(y=a x^k\), because log y against log x is the straighter graph

4. (d) The line of best fit for \(\log_{10}y\) against \(\log_{10}x\) has gradient 0.748 and intercept 0.553. Write the model in the form \(y=a x^k\). [3]

   • M1 for k = gradient = 0.748
   • M1 for a = 10^intercept = 10^0.553
   • A1 y = 3.57x^0.748

5. (e) Use the model to estimate y when x = 40. [2]

   • M1 for substituting x = 40
   • A1 56.4

6. (f) A different relationship has the form \(y=a\times b^{x}\). The graph of \(\log_{10}y\) against x is a straight line through \((0, 0.9)\) and \((5, 2.4)\). Find a and b. [3]

   • M1 for gradient = (2.4 − 0.9) ÷ (5 − 0) = 0.300
   • M1 for b = 10^gradient and a = 10^intercept (intercept 0.900)
   • A1 a = 7.94, b = 2.00

The mass M grams of a radioactive sample, t days after it is first measured, is modelled by \(M=250\times0.92^{t}\). [7]

1. (a) Write down the value of M when \(t=0\), and find the value of M when \(t=10\). Give your answer to 3 significant figures. [2]

   • A1 250
   • A1 109 (\(250\times0.92^{10}\))

2. (b) Write down the percentage of the M lost each day. [1]

   • A1 8% (0.92 means a decrease of 8%)

3. (c) Use your calculator to find when \(M=100\). Give your answer to 3 significant figures. [2]

   • M1 for solving \(250\times0.92^{t}=100\)
   • A1 t = 11.0

4. (d) The sample is safe to handle once it has less than 20 g. Find the first whole number of days after which it is safe. [2]

   • M1 for solving \(250\times0.92^{t}=20\) (t = 30.3)
   • A1 31 (the first whole number above 30.3)

A culture of bacteria grows so that the number of bacteria N after t hours is modelled by \(N=N_0e^{kt}\), where \(N_0\) and k are constants. At \(t=0\) there are 1200 bacteria. At \(t=4\) there are 3100 bacteria. [8]

1. (a) Find the value of k. Give your answer to 3 significant figures. [3]

   • M1 for \(1200e^{4k}=3100\), so \(e^{4k}=\dfrac{31}{12}\)
   • M1 for \(4k=\ln\dfrac{31}{12}\)
   • A1 k = 0.237

2. (b) Find the time at which there are 20 000 bacteria. Give your answer to 3 significant figures. [2]

   • M1 for \(1200e^{kt}=20\,000\), so \(e^{kt}=\dfrac{50}{3}\), and \(t=\dfrac{\ln(50/3)}{k}\)
   • A1 t = 11.9

3. (c) At \(t=12\) an antibiotic is added. From then the number of bacteria is modelled by \(N=20\,000\,e^{-0.35s}\), where s is the time in hours since the antibiotic was added. Find the half-life of this decay. Give your answer to 3 significant figures. [3]

   • M1 for \(20\,000e^{-0.35s}=10\,000\), so \(e^{-0.35s}=0.5\)
   • M1 for \(-0.35s=\ln0.5=-\ln2\), so \(s=\dfrac{\ln2}{0.35}\)
   • A1 1.98 hours

The height H cm of a seedling d days after planting is modelled by \(H=a+b\ln d\), where a and b are constants. The seedling is 3.2 cm tall after 2 days and 9.8 cm tall after 14 days. [8]

1. (a) Find the value of b and the value of a. Give your answers to 3 significant figures. [4]

   • M1 for \(a+b\ln2=3.2\) and \(a+b\ln14=9.8\)
   • M1 for subtracting: \(b(\ln14-\ln2)=6.6\), so \(b=\dfrac{6.6}{\ln7}\)
   • A1 b = 3.39
   • A1 a = 0.849 (using the full value of b)

2. (b) Use the model to predict the height of the seedling after 30 days. Give your answer to 3 significant figures. [2]

   • M1 for \(H=a+b\ln30\)
   • A1 12.4 cm

3. (c) Use the model to find the number of days after planting when the seedling is 15 cm tall. Give your answer to 3 significant figures. [2]

   • M1 for solving \(a+b\ln d=15\)
   • A1 64.9 days