Exponents and logarithms

Worked example 1 | Count the factors

Question Simplify \(5^3\times5^2\).

Solution

Using \(a^m a^n=a^{m+n}\), \[5^3\times5^2=(5\times5\times5)(5\times5)=5^{3+2}=5^5.\] The base stays 5; there are five factors of 5 altogether.

Worked example 2 | Cancel common factors

Question Simplify \(\dfrac{6^4}{6^3}\).

Solution

Using \(a^m/a^n=a^{m-n}\), \[\frac{6^4}{6^3}=\frac{6\times6\times6\times6}{6\times6\times6}=6^{4-3}=6.\]

Worked example 3 | What is raised to the power zero?

Question Evaluate

1. \(7^0\)

2. \((-4)^0\)

3. \(5x^0\), where \(x\ne0\).

Solution

Every non-zero base to the power zero is 1. (a) \(7^0=1,\) (b) \((-4)^0=1,\) (c) \(5x^0=5\times1=5.\) In (c), only \(x\) is raised to the power zero. By contrast, \((5x)^0=1\).

Worked example 4 | Rewrite, then evaluate

Question Evaluate

1. \(2^{-3}\)

2. \(\left(\frac23\right)^{-2}\)

3. \((-2)^{-3}\).

Solution

Apply \(a^{-n}=1/a^n\) to the whole base. (a) \(2^{-3}=\frac1{2^3}=\frac18,\) (b) \(\left(\frac23\right)^{-2}=\frac1{(2/3)^2}=\frac1{4/9}=\frac94,\) (c) \((-2)^{-3}=\frac1{(-2)^3}=\frac1{-8}=-\frac18.\)

Worked example 5 | Combining negative exponents

Question Simplify

1. \(5^3\times5^{-6}\)

2. \(\frac{x^{-2}}{x^{-5}}\), where \(x\ne0\).

Solution

Add exponents when multiplying; subtract them when dividing. (a) \(5^{3+(-6)}=5^{-3}=\frac1{125},\) (b) \(x^{-2-(-5)}=x^3.\)

Worked example 6 | Multiply, do not add

Question Simplify \((2^3)^{-4}\).

Solution

Using \((a^m)^n=a^{mn}\), \[(2^3)^{-4}=2^{3\times(-4)}=2^{-12}=\frac1{2^{12}}=\frac1{4096}.\]

Worked example 7 | Include the numerical coefficient

Question Simplify

1. \((2x)^4\)

2. \(\left(\dfrac{3x}{2}\right)^2\).

Solution

Raise each factor, and each part of the quotient, to the stated power. (a) \((2x)^4=2^4x^4=16x^4,\) (b) \(\left(\frac{3x}{2}\right)^2=\frac{3^2x^2}{2^2}=\frac{9x^2}{4}.\)

Worked example 8 | Use several rules in sequence

Question Simplify \(\dfrac{(2x^{-3}y^2)^2}{4x^{-1}y^{-3}}\), giving your answer with positive exponents. Assume \(x,y\ne0\).

Solution

First expand the power using \((ab)^n=a^nb^n\) and \((a^m)^n=a^{mn}\). Then use \(a^m/a^n=a^{m-n}\). \[\begin{aligned} \frac{(2x^{-3}y^2)^2}{4x^{-1}y^{-3}} &=\frac{4x^{-6}y^4}{4x^{-1}y^{-3}}\\[6pt] &=x^{-6-(-1)}y^{4-(-3)}\\[6pt] &=x^{-5}y^7=\boxed{\frac{y^7}{x^5}}. \end{aligned}\] Keep the original restrictions \(x,y\ne0\), even after simplifying.

Worked example 9 | Translate in both directions

Question

1. Write \(2^5=32\) in logarithmic form.

2. Write \(\log_{10}0.01=-2\) in exponential form.

Solution

Use \(a^x=b\Longleftrightarrow\log_a b=x\); keep the same base. (a) \(\boxed{\log_2 32=5},\) (b) \(\boxed{10^{-2}=0.01}.\) In (a), the logarithm is 5, not 32: 5 is the required exponent.

Worked example 10 | Exact logarithms without a calculator

Question Evaluate

1. \(\log1000\)

2. \(\log0.001\)

3. \(\log1\).

Solution

Ask which power of 10 produces each number. (a) \(10^3=1000\quad\) \(\Rightarrow\quad \quad \log1000=3,\) (b) \(10^{-3}=0.001\quad\) \(\Rightarrow\quad \quad \log0.001=-3,\) (c) \(10^0=1\quad\) \(\Rightarrow\quad \quad \log1=0.\)

Worked example 11 | Recognise powers of \(e\)

Question Evaluate

1. \(\ln e\)

2. \(\ln1\)

3. \(\ln(e^4)\)

4. \(\ln(1/e^2)\).

Solution

Identify the exponent on \(e\) in each case. (a) \(e^1=e\quad\) \(\Rightarrow\quad \quad \ln e=1,\) (b) \(e^0=1\quad\) \(\Rightarrow\quad \quad \ln1=0,\) (c) \(e^4=e^4\quad\) \(\Rightarrow\quad \quad \ln(e^4)=4,\) (d) \(\frac1{e^2}=e^{-2}\quad\) \(\Rightarrow\quad \quad \ln(1/e^2)=-2.\) These are exact answers; no decimal approximation to \(e\) is needed.

Worked example 12 | Translate and interpret

Question

1. Express \(e^x=7\) in logarithmic form.

2. Express \(\ln y=-2\) in exponential form and find the exact value of \(y\).

Solution

Use \(e^x=b\Longleftrightarrow x=\ln b\). (a) \(\boxed{x=\ln7},\) (b) \(\boxed{y=e^{-2}=\frac1{e^2}}.\) In (b), \(y\) is positive even though its natural logarithm is negative.

Worked example 13 | Calculate and check the size [GDC]

Question Evaluate to 3 significant figures:

1. \(\log50\);

2. \(\ln7\);

3. \(\ln0.2\).

Solution

Enter each complete expression in Scratchpad Calculate.

Since \(10^1<50<10^2\), \(\log50\) must lie between 1 and 2. Since \(0<0.2<1\), its natural logarithm must be negative.

Worked example 14 | A logarithm between integer powers [GDC]

Question The number \(x\) satisfies \(10^x=350\). Write \(x\) exactly, then give its value to 3 significant figures.

Solution

Use the given conversion rule \(a^x=b\Longleftrightarrow x=\log_a b\): \[x=\log_{10}350=\log350\quad\text{(exact)}.\] Enter log(350) to obtain \(2.544068044\ldots\), so \(\boxed{x\approx2.54}\).

The answer is sensible because \(10^2<350<10^3\). To check numerically, enter 10^(log(350)); it should return 350 (up to numerical rounding).