Mixed exponent and logarithm practice Problem set B One question at a time

Given that \(x=\log_{10}a\), \(y=\log_{10}b\) and \(z=\log_{10}c\), write in terms of x, y and z:

1. (a) \(\log_{10}(a^2c^3)\)

   • \(\log_{10}(a^2c^3)=2\log_{10}a+3\log_{10}c\)
   • \(=2x+3z\)

2. (b) \(\log_{10}\!\left(\dfrac{a^3}{bc^2}\right)\)

   • \(\log_{10}\dfrac{a^3}{bc^2}=3\log_{10}a-\log_{10}b-2\log_{10}c\)
   • \(=3x-y-2z\)

3. (c) \(\log_{10}\!\left(\dfrac{10c^2}{\sqrt a}\right)\)

   • \(\log_{10}10=1\) and \(\log_{10}\sqrt a=\tfrac12x\)
   • \(\log_{10}\dfrac{10c^2}{\sqrt a}=1+2z-\tfrac12x\)

Write each expression as a single logarithm.

1. (a) \(5\ln p+3\ln q\)

   • \(5\ln p=\ln p^5\) and \(3\ln q=\ln q^3\)
   • \(\ln p^5+\ln q^3=\ln(p^5q^3)\)

2. (b) \(\tfrac14\ln m-2\ln n\)

   • \(\tfrac14\ln m=\ln m^{1/4}\) and \(2\ln n=\ln n^2\)
   • \(\ln m^{1/4}-\ln n^2=\ln\dfrac{\sqrt[4]{m}}{n^2}\)

3. (c) \(\ln20-\ln4+3\ln2\)

   • \(\ln20-\ln4=\ln5\) and \(3\ln2=\ln8\)
   • \(\ln5+\ln8=\ln40\)

Solve each equation.

1. (a) \(\log_{10}(x-7)=3\)

   • \(x-7=10^3=1000\)
   • x = 1007

2. (b) \(\log_{10}(4x+1)=2\)

   • \(4x+1=10^2=100\)
   • 4x = 99, so x = 24.75

3. (c) \(\ln(x+2)=1.5\)

   • \(x+2=e^{1.5}\)
   • x = e^1.5 − 2 = 2.48

Use logarithms to solve each equation. Give x to 3 significant figures.

1. (a) \(6^x=45\)

   • \(x\ln6=\ln45\)
   • x = ln 45 / ln 6 = 2.12

2. (b) \(4\times3^x+5=80\)

   • \(4\times3^x=75\), so \(3^x=18.75\)
   • \(x\ln3=\ln18.75\)
   • x = 2.67

3. (c) \(1.8\times1.05^x=36\)

   • \(1.05^x=\dfrac{36}{1.8}=20\)
   • \(x\ln1.05=\ln20\)
   • x = 61.4

Find the exact value of each logarithm.

1. (a) \(\log_{10}\dfrac{1}{\sqrt[4]{10}}\)

   • \(\dfrac1{\sqrt[4]{10}}=10^{-1/4}\)
   • \(\log_{10}10^{-1/4}=-\tfrac14\)

2. (b) \(\log_{10}\sqrt{10^{7}}\)

   • \(\sqrt{10^7}=10^{7/2}\)
   • \(\log_{10}10^{7/2}=\tfrac72\)

3. (c) \(\ln\dfrac{e^{4}}{\sqrt[3]{e}}\)

   • \(\dfrac{e^4}{\sqrt[3]{e}}=e^{4-1/3}=e^{11/3}\)
   • \(\ln e^{11/3}=\tfrac{11}3\)

Solve each equation for x, where \(x>0\) and \(x\ne1\).

1. (a) \(\log_{x}125=3\)

   • \(x^3=125\)
   • \(x=5\)

2. (b) \(\log_{x}64=-3\)

   • \(x^{-3}=64\)
   • \(x^3=\tfrac1{64}\), so \(x=\tfrac14\)

3. (c) \(\log_{x}27=\tfrac32\)

   • \(x^{3/2}=27\)
   • \(x=27^{2/3}=(3^3)^{2/3}=9\)

Solve each equation by writing both sides as powers of the same base.

1. (a) \(8^{x+1}=16^{2x-3}\)

   • \(8^{x+1}=2^{3x+3}\) and \(16^{2x-3}=2^{8x-12}\)
   • 3x + 3 = 8x − 12
   • x = 3

2. (b) \(25^{2x-1}=125^{x+2}\)

   • \(25^{2x-1}=5^{4x-2}\) and \(125^{x+2}=5^{3x+6}\)
   • 4x − 2 = 3x + 6
   • x = 8

3. (c) \(2^{3x+1}=\dfrac1{64}\)

   • \(\tfrac1{64}=2^{-6}\)
   • 3x + 1 = −6
   • \(x=-\tfrac73\)

Solve \(5^{2x-1}=3^{x+4}\), giving x in the form \(\dfrac{\ln p}{\ln q}\). Then give x to 3 significant figures.

• \((2x-1)\ln5=(x+4)\ln3\)
• \(2x\ln5-\ln5=x\ln3+4\ln3\)
• \(x(2\ln5-\ln3)=4\ln3+\ln5=\ln405\)
• \(2\ln5-\ln3=\ln\tfrac{25}3\), so \(x=\dfrac{\ln405}{\ln(25/3)}\)=2.83

Solve each equation.

1. (a) \(\log_{10}(8x) = 2\)

   • \(8x = 10^2 = 100\)
   • x = 12.5

2. (b) \(\log_{10}(x-3) = -2\)

   • \(x - 3 = 10^{-2} = 0.01\)
   • x = 3.01

3. (c) \(\ln(4x+3) = 2\)

   • \(4x + 3 = e^{2}\)
   • \(x = \frac{e^{2} - 3}{4} = 1.10\)

Solve each equation by writing both sides as powers of the same base.

1. (a) \(7^{x-1} = 343\)

   • \(343 = 7^3\), so \(x - 1 = 3\)
   • x = 4

2. (b) \(4^{x} = 32\)

   • \(4 = 2^2\) and \(32 = 2^5\), so \(2^{2x} = 2^5\)
   • \(2x = 5\), so \(x = \frac{5}{2}\)

3. (c) \(3^{2x-1} = 81^{x-3}\)

   • \(81 = 3^4\), so \(81^{x-3} = 3^{4(x-3)}\)
   • \(2x - 1 = 4x - 12\)
   • \(2x = 11\), so \(x = \frac{11}{2}\)

The value of a car is modelled by \(V=24\,000\times0.88^{t}\) euros, where t is the time in years after it is bought.

1. (a) Find the value when it is bought.

   • t = 0: \(V=24\,000\times0.88^{0}=24\,000\)

2. (b) Find the value after 3 years.

   • \(V=24\,000\times0.88^{3}\)
   • From the calculator: 16 355.33 euros

3. (c) Use your calculator to find when the value has fallen to 10 000 euros.

   • \(24\,000\times0.88^{t}=10\,000\)
   • Solve for t: t = 6.85 years

The number of users, in thousands, of two apps after t months is \(P=300\times1.15^{t}\) and \(Q=700\times1.06^{t}\). Find when the two numbers are equal. Give your answer to 3 significant figures.

• \(300\times1.15^{t}=700\times1.06^{t}\)
• \(\left(\dfrac{1.15}{1.06}\right)^{t}=\dfrac{700}{300}=\dfrac73\)
• \(t=\dfrac{\ln(7/3)}{\ln(1.15/1.06)}\)
• t = 10.4 months

The number of views of a video doubles every 4 hours. Find how long it takes to multiply by 30. Give your answer in hours to 3 significant figures.

• After t hours the number is multiplied by \(2^{t/4}\)
• \(2^{t/4}=30\)
• \(\dfrac{t}{4}=\dfrac{\ln30}{\ln2}\)
• t = 19.6 hours

The number of members of an online forum is modelled by \(N=320\times1.35^{t}\), where t is the time in weeks.

1. (a) Write down the number of members at the start, and find the number after 6 weeks. Give your answer to 3 significant figures.

   • t = 0: \(N=320\times1.35^{0}=320\)
   • \(N=320\times1.35^{6}\)
   • From the calculator: 1940

2. (b) Use your calculator to find when there are 5000 members. Give t to 3 significant figures.

   • \(320\times1.35^{t}=5000\)
   • Solve for t: t = 9.16 weeks

3. (c) Write down the percentage increase each week.

   • Each week the number is multiplied by 1.35.
   • 1.35 means an increase of 35%.

The amount of caffeine in a person's blood is modelled by \(C=C_0e^{-0.139t}\), where t is the time in hours after drinking a coffee.

1. (a) Find the half-life of caffeine in the blood. Give your answer to 3 significant figures.

   • \(e^{-0.139t}=0.5\)
   • \(t=\dfrac{\ln2}{0.139}=4.99\) hours

2. (b) Find the time for the amount to fall to 15% of its starting value. Give your answer to 3 significant figures.

   • \(e^{-0.139t}=0.15\)
   • \(t=\dfrac{\ln(1/0.15)}{0.139}=13.6\) hours

The number of views of a video is modelled by \(V=V_0e^{kt}\), where t is the time in days. There are 2500 views at \(t=0\) and 9000 views at \(t=5\).

1. (a) Find k to 3 significant figures.

   • \(2500e^{5k}=9000\), so \(e^{5k}=3.6\)
   • \(k=\dfrac{\ln3.6}{5}=0.256\)

2. (b) Find when there are 50 000 views. Give your answer to 3 significant figures.

   • \(2500e^{kt}=50\,000\), so \(e^{kt}=20\)
   • \(t=\dfrac{\ln20}{k}=11.7\) days

Given that \(9\times6^{x}=24\times3^{x}\):

1. (a) write down the exact value of \(2^x\);

   • \(6^x=2^x\times3^x\)
   • \(9\times2^x\times3^x=24\times3^x\)
   • \(2^x=\dfrac{24}{9}=\dfrac83\)

2. (b) hence find x.

   • \(x=\dfrac{\ln(8/3)}{\ln2}\)
   • x = 1.42

Solve each pair of simultaneous equations.

1. (a) \(\log_{10}x+\log_{10}y=5\) and \(\log_{10}x-4\log_{10}y=0\)

   • Let \(u=\log_{10}x\), \(v=\log_{10}y\): \(u+v=5\) and \(u-4v=0\)
   • Subtract: 5v = 5, so v = 1 and u = 4
   • \(x=10^4=10\,000\), \(y=10\)

2. (b) \(3\log_{10}x-\ln y=7\) and \(\log_{10}x^{2}+3\ln y=12\)

   • Let \(u=\log_{10}x\), \(v=\ln y\): \(3u-v=7\) and \(2u+3v=12\)
   • From the first, v = 3u − 7. Then 2u + 9u − 21 = 12, so u = 3 and v = 2
   • \(x=10^3=1000\), \(y=e^2\)

In each part a and b are constants with \(a>1\) and \(b>0\). Find the product of all the solutions for x.

1. (a) \(\log_a(x^6)=b\)

   • \(x^6=a^b\), so \(x=a^{b/6}\) or \(x=-a^{b/6}\)
   • Product: \(-a^{b/6}\times a^{b/6}=-a^{b/3}\)

2. (b) \((\log_ax)^2-7\log_ax+10=0\)

   • Let \(L=\log_ax\): \(L^2-7L+10=0\), so \(L=2\) or \(L=5\)
   • \(x=a^2\) or \(x=a^5\)
   • Product: \(a^7\)

Solve \(8^{x} = \dfrac{4^{x+3}}{2}\).

• \(8 = 2^3\), \(4 = 2^2\) and \(2 = 2^1\)
• \(2^{3x} = \dfrac{2^{2(x+3)}}{2^1} = 2^{2x+5}\)
• \(3x = 2x + 5\), so x = 5