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Logarithms

Definition

Evaluating Logarithms

Exercise
Evaluate:
\(\log 100=\)

$$\begin{aligned}[t]\log(100) &= \log\left(10^{2}\right) \\ &=2\\ \end{aligned}$$

Exercise
Evaluate:
\(\log 0.1=\)

$$\begin{aligned}[t]\log(0.1) &= \log\left(10^{-1}\right) \\ &= -1 \\ \end{aligned}$$

Exercise
Evaluate:
\(\log \left(\dfrac{1}{100}\right) =\)

$$\begin{aligned}[t]\log\left(\dfrac{1}{100}\right) &= \log\left(\dfrac{1}{10^2}\right) \\ &= \log\left(10^{-2}\right) \\ &= -2 \\ \end{aligned}$$

Exercise
Evaluate:
\(\log \sqrt{10} =\)

$$\begin{aligned}[t]\log(\sqrt{10}) &= \log(10^{0.5}) \\ &= 0.5 \\ \end{aligned}$$

Exercise
Evaluate:
\(\log 1=\)

$$\begin{aligned}[t]\log(1) &= \log\left(10^{0}\right) \\ &=0\\ \end{aligned}$$

Evaluating using a calculator

Exercise

Evaluate (round to 2 decimal places).
\(\log (2) \approx\)

By entering \(\log(2)\) and pressing the equal button, the calculator displays: \(0.30103\).
So, \(\log(2) \approx 0.30\) (rounded to two decimal places).

Exercise

Evaluate (round to 2 decimal places).
\(\log (0.2) \approx\)

By entering \(\log(0.2)\) and pressing the equal button, the calculator displays: \(-0.69897\).
So, \(\log(0.2) \approx -0.70\) (rounded to two decimal places).

Exercise

Evaluate (round to 2 decimal places).
\(\log(2 \times 10^9) \approx\)

By entering \(\log(2 \times 10^9)\) and pressing the equal button, the calculator displays: \(9.30103\).
So, \(\log(2 \times 10^9) \approx 9.30\) (rounded to two decimal places).

Solving Exponential Equations Using Logarithms

Exercise

Find \(x\) such that \(8 = 10^{x}\).
\(x \approx\) (rounded to 3 decimal places)

To solve \(8 = 10^{x}\), take the logarithm (base \(10\)) of both sides:$$\begin{aligned}8 &= 10^x \\ \log(8) &= \log(10^x) \\ \log(8) &= x \\ x &\approx 0.903\end{aligned}$$So, \(x \approx 0.903\) (rounded to 3 decimal places).

Exercise

Find \(x\) such that \(0.4 = 10^{x}\).
\(x \approx\) (rounded to 3 decimal places)

To solve \(0.4 = 10^{x}\), take the logarithm (base \(10\)) of both sides:$$\begin{aligned}0.4 &= 10^x \\ \log(0.4) &= \log(10^x) \\ \log(0.4) &= x \\ x &\approx -0.398\end{aligned}$$So, \(x \approx -0.398\) (rounded to 3 decimal places).

Exercise

Find \(x\) such that \(250 = 10^{x}\).
\(x \approx\) (rounded to 3 decimal places)

To solve \(250 = 10^{x}\), take the logarithm (base \(10\)) of both sides:$$\begin{aligned}250 &= 10^x \\ \log(250) &= \log(10^x) \\ \log(250) &= x \\ x &\approx 2.398\end{aligned}$$So, \(x \approx 2.398\) (rounded to 3 decimal places).

Solving for \(x\) When \(\log(x)\) Is Given

Exercise

Find \(x\) such that \(\log(x) = 3\).
\(x =\)

Take \(10^{\cdot}\) on both sides:$$\begin{aligned}\log(x) &= 3 \\ 10^{\log(x)} &= 10^{3} \\ x &= 10^{3} \\ x &= 1000\end{aligned}$$

Exercise

Find \(x\) such that \(\log(x) = -1\).
\(x =\)

Take \(10^{\cdot}\) on both sides:$$\begin{aligned}\log(x) &= -1 \\ 10^{\log(x)} &= 10^{-1} \\ x &= 10^{-1} \\ x &= 0.1\end{aligned}$$

Exercise

Find \(x\) such that \(\log(x) = 0\).
\(x =\)

Take \(10^{\cdot}\) on both sides:$$\begin{aligned}\log(x) &= 0 \\ 10^{\log(x)} &= 10^{0} \\ x &= 10^{0} \\ x &= 1\end{aligned}$$

Exercise

Find \(x\) such that \(\log(x) = 7\).
\(x =\)

Take \(10^{\cdot}\) on both sides:$$\begin{aligned}\log(x) &= 7 \\ 10^{\log(x)} &= 10^{7} \\ x &= 10^{7} \\ x &= 10\,000\,000\end{aligned}$$

Laws of Logarithms

Writing as a Single Logarithm: Level 1

Exercise
Write as a single logarithm
\(\log (5)+\log (3)=\)

$$\begin{aligned}[t]\log (5)+\log (3) &= \log (5\times 3) \\ &= \log 15\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\) :
\(\log (15) - \log (5) =\)

$$\begin{aligned}[t]\log (15) - \log (5) &= \log \left(\dfrac{15}{5}\right) \\ &= \log(3)\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(\log (4) + \log \left(\dfrac{1}{2}\right) = \)

$$\begin{aligned}[t]\log(4) + \log\left(\dfrac{1}{2}\right) &= \log \left(4 \times \dfrac{1}{2}\right) \\ &= \log(2)\end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(\log (18) - \log (3) =\)

$$\begin{aligned}[t]\log(18) - \log(3) &= \log\left(\dfrac{18}{3}\right) \\ &= \log(6)\end{aligned}$$

Writing as a Single Logarithm: Level 2

Exercise
Write as a single logarithm in the form \(\log k\):
\(\log(8) + 1 = \)

$$\begin{aligned}[t]\log(8) + 1 &= \log(8) + \log(10) \\ &= \log(8 \times 10) \\ &= \log(80)\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(\log(3) + 2 = \)

$$\begin{aligned}[t]\log(3) + 2 &= \log(3) + \log(10^2) \\ &= \log(3) + \log(100) \\ &= \log(3 \times 100) \\ &= \log(300)\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(2 - \log(25) = \)

$$\begin{aligned}[t]2 - \log(25) &= \log(10^2) - \log(25) \\ &= \log\left(\frac{100}{25}\right) \\ &= \log(4)\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(\log(200) - 2 = \)

$$\begin{aligned}[t]\log(200) - 2 &= \log(200) - \log(10^2) \\ &= \log\left(\dfrac{200}{100}\right) \\ &= \log(2)\\ \end{aligned}$$

Writing as a Single Logarithm: Level 3

Exercise
Write as a single logarithm in the form \(\log k\):
\(2\log(3) + 1 = \)

$$\begin{aligned}[t]2\log(3) + 1 &= \log(3^2) + \log(10^1) \\ &= \log(9) + \log(10) \\ &= \log(9 \times 10) \\ &= \log(90)\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(3\log(2) - \log(4) = \)

$$\begin{aligned}[t]3\log(2) - \log(4) &= \log(2^3) - \log(4) \\ &= \log(8) - \log(4) \\ &= \log\left(\dfrac{8}{4}\right) \\ &= \log(2)\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(2\log(20) - 2 = \)

$$\begin{aligned}[t]2\log(20) - 2 &= \log(20^2) - \log(10^2) \\ &= \log(400) - \log(100) \\ &= \log\left(\dfrac{400}{100}\right) \\ &= \log(4)\\ \end{aligned}$$

Exercise
Write as a single logarithm in the form \(\log k\):
\(2\log(30) - 1 = \)

$$\begin{aligned}[t]2\log(30) - 1 &= \log(30^2) - \log(10^1) \\ &= \log(900) - \log(10) \\ &= \log\left(\dfrac{900}{10}\right) \\ &= \log(90)\\ \end{aligned}$$

Using Logarithms to Solve Exponential Equations

Solving Exponential Equations: Level 1

Exercise

Solve \(2^x = 7\) (give your answer to 3 decimal places).
\(x=\)

$$\begin{aligned}2^x &= 7 \\ \log(2^x) &= \log 7 &&\text{(taking log of both sides)}\\ x \log 2 &= \log 7 &&\text{(power rule)}\\ x &= \frac{\log 7}{\log 2} &&\text{(dividing both sides by } \log 2\text{)}\\ x &\approx 2.807 &&\text{(using calculator)}\\ \end{aligned}$$

Exercise

Solve \(3^x = 15\) (give your answer to 3 decimal places).
\(x=\)

$$\begin{aligned}3^x &= 15 \\ \log(3^x) &= \log 15 &&\text{(taking log of both sides)}\\ x \log 3 &= \log 15 &&\text{(power rule)}\\ x &= \frac{\log 15}{\log 3} &&\text{(dividing both sides by } \log 3\text{)}\\ x &\approx 2.465 &&\text{(using calculator)}\\ \end{aligned}$$

Exercise

Solve \(5^x = 100\) (give your answer to 3 decimal places).
\(x=\)

$$\begin{aligned}5^x &= 100 \\ \log(5^x) &= \log 100 &&\text{(taking log of both sides)}\\ x \log 5 &= \log 100 &&\text{(power rule)}\\ x &= \frac{\log 100}{\log 5} &&\text{(dividing both sides by } \log 5\text{)}\\ x &\approx 2.861 &&\text{(using calculator)}\\ \end{aligned}$$

Exercise

Solve \(6^x = 80\) (give your answer to 3 decimal places).
\(x=\)

$$\begin{aligned}6^x &= 80 \\ \log(6^x) &= \log 80 &&\text{(taking log of both sides)}\\ x \log 6 &= \log 80 &&\text{(power rule)}\\ x &= \frac{\log 80}{\log 6} &&\text{(dividing both sides by } \log 6\text{)}\\ x &\approx 2.446 &&\text{(using calculator)}\\ \end{aligned}$$

Solving Exponential Equations: Level 2

Exercise

Solve \(5 \cdot 2^x = 7\) (give your answer to 3 decimal places).
\(x=\)

$$\begin{aligned}5 \cdot 2^x &= 7 \\ 2^x &= \frac{7}{5} &&\text{(dividing both sides by 5)}\\ \log(2^x) &= \log \left(\frac{7}{5}\right) &&\text{(taking log of both sides)}\\ x \log 2 &= \log \left(\frac{7}{5}\right) &&\text{(power rule)}\\ x &= \frac{\log \left(\frac{7}{5}\right)}{\log 2} &&\text{(dividing both sides by } \log 2\text{)}\\ x &\approx 0.485 &&\text{(using calculator)}\end{aligned}$$

Exercise

Solve \(-\left(2^x\right) = -10\) (give your answer to 3 decimal places).
\(x=\)

$$\begin{aligned}-\left(2^x\right) &= -10 \\ 2^x &= 10 &&\text{(dividing both sides by \(-1\))}\\ \log(2^x) &= \log 10 &&\text{(taking log of both sides)}\\ x \log 2 &= \log 10 &&\text{(power rule)}\\ x &= \frac{\log 10}{\log 2} &&\text{(dividing both sides by } \log 2\text{)}\\ x &\approx 3.322 &&\text{(using calculator)}\end{aligned}$$

Exercise

Solve \(4 \cdot 3^x = 60\) (give your answer to 3 decimal places).
\(x=\)

$$\begin{aligned}4 \cdot 3^x &= 60 \\ 3^x &= \frac{60}{4} &&\text{(dividing both sides by 4)}\\ 3^x &= 15 \\ \log(3^x) &= \log 15 &&\text{(taking log of both sides)}\\ x \log 3 &= \log 15 &&\text{(power rule)}\\ x &= \frac{\log 15}{\log 3} &&\text{(dividing both sides by } \log 3\text{)}\\ x &\approx 2.465 &&\text{(using calculator)}\end{aligned}$$

Exercise

Solve \(-2 \cdot (0.5)^x = -4\).
\(x=\)

$$\begin{aligned}-2 \cdot (0.5)^x &= -4 \\ (0.5)^x &= \frac{-4}{-2} &&\text{(dividing both sides by \(-2\))}\\ (0.5)^x &= 2 \\ \log\left((0.5)^x\right) &= \log(2) \\ x \cdot \log(0.5) &= \log(2) \\ x &= \frac{\log(2)}{\log(0.5)} \\ x &= -1\end{aligned}$$So, \(x = -1\).

Applications of Logarithms

Applying Logarithms in Science

Exercise

The pH scale in chemistry is \(\text{pH} = -\log_{10} [H^+]\) where \([H^+]\) is the hydrogen ion concentration in moles per litre.
The pH of a solution is \(3.2\). Find the hydrogen ion concentration \([H^+]\) (give your answer in scientific notation with 3 significant digits).
\(\times\) mol/L

We know:$$\begin{aligned}\text{pH} &= -\log_{10}[H^+] \\ 3.2 &= -\log_{10}[H^+] &&\text{(substituting the value)}\\ -3.2 &= \log_{10}[H^+] &&\text{(multiplying both sides by \(-1\))}\\ 10^{-3.2} &= 10^{\log_{10}[H^+]} &&\text{(exponentiating both sides)}\\ 10^{-3.2}&= [H^+] &&(10^{\log_{10}x}=x)\\ [H^+] &\approx 0.00063096 &&\text{(using calculator)}\\ [H^+] &\approx 6.31 \times 10^{-4}\ \text{mol/L} &&{\scriptstyle\text{(in scientific notation with 3 significant digits)}}\\ \end{aligned}$$

Exercise

The Richter scale measures earthquake intensity using the formula \(M = \log_{10} \left(\frac{I}{I_0}\right)\), where \(M\) is the magnitude, \(I\) is the intensity of the earthquake, and \(I_0\) is the intensity of a standard earthquake.
An earthquake has a magnitude of \(4.5\) on the Richter scale. Find the intensity ratio \(\frac{I}{I_0}\) (give your answer in scientific notation with 3 significant digits).
\(\frac{I}{I_0}=\)\(\times\)

We know:$$\begin{aligned}M &= \log_{10} \left(\frac{I}{I_0}\right) \\ 4.5 &= \log_{10} \left(\frac{I}{I_0}\right) &&\text{(substituting the value)}\\ 10^{4.5} &= 10^{\log_{10} \left(\frac{I}{I_0}\right)} &&\text{(exponentiating both sides)}\\ 10^{4.5} &= \frac{I}{I_0} &&(10^{\log_{10}x}=x)\\ \frac{I}{I_0} &\approx 31622.7766 &&\text{(using calculator)}\\ \frac{I}{I_0} &\approx 3.16 \times 10^{4} &&{\scriptstyle\text{(in scientific notation with 3 significant digits)}}\\ \end{aligned}$$

Exercise

The intensity of sound is measured in decibels (dB) using the formula \(L = 10 \log_{10} \left(\frac{I}{I_0}\right)\), where \(L\) is the sound level in decibels, \(I\) is the intensity of the sound, and \(I_0\) is the reference intensity (threshold of human hearing).
A sound has a level of \(75\) decibels. Find the intensity ratio \(\frac{I}{I_0}\) (give your answer in scientific notation with 3 significant digits).
\(\frac{I}{I_0}=\)\(\times\)

We know:$$\begin{aligned}L &= 10 \log_{10} \left(\frac{I}{I_0}\right) \\ 75 &= 10 \log_{10} \left(\frac{I}{I_0}\right) &&\text{(substituting the value)}\\ 7.5 &= \log_{10} \left(\frac{I}{I_0}\right) &&\text{(dividing both sides by 10)}\\ 10^{7.5} &= 10^{\log_{10} \left(\frac{I}{I_0}\right)} &&\text{(exponentiating both sides)}\\ 10^{7.5} &= \frac{I}{I_0} &&(10^{\log_{10}x}=x)\\ \frac{I}{I_0} &\approx 3162277.66 &&\text{(using calculator)}\\ \frac{I}{I_0} &\approx 3.16 \times 10^{7} &&{\scriptstyle\text{(in scientific notation with 3 significant digits)}}\\ \end{aligned}$$