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. \[\begin{aligned} \text{(a)}\quad 7^0&=1,\\ \text{(b)}\quad (-4)^0&=1,\\ \text{(c)}\quad 5x^0&=5\times1=5. \end{aligned}\] In (c), only \(x\) is raised to the power zero: read \(5x^0\) as \(5\times(x^0)\). By contrast, \((5x)^0=1\), because the whole non-zero product \(5x\) is the base.

Worked example 4 | Rewrite, then evaluate

Question:  Evaluate

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

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

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

Solution

First use \(a^{-n}=1/a^n\). Then calculate the positive power.

(a)  \(\displaystyle 2^{-3}=\frac{1}{2^3}=\frac{1}{2\times2\times2}=\boxed{\frac18}\).

(b)  \(\displaystyle \left(\frac23\right)^{-2}=\frac{1}{(2/3)^2} =1\div\frac49=1\times\frac94=\boxed{\frac94}\).

Dividing by a fraction means multiplying by its reciprocal. Equivalently, invert \(2/3\) first and calculate \((3/2)^2\).

(c)  \(\displaystyle (-2)^{-3}=\frac{1}{(-2)^3}=\frac{1}{(-2)(-2)(-2)}=\frac1{-8}=\boxed{-\frac18}\).

Worked example 5 | Combining negative exponents

Question:  Simplify

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

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

Solution

Choose the rule from the operation between the powers.

(a)  Multiplication: use \(a^m a^n=a^{m+n}\). \[5^3\times5^{-6}=5^{3+(-6)}=5^{3-6}=5^{-3}=\frac1{5^3}=\boxed{\frac1{125}}.\]

(b)  Division: use \(a^m/a^n=a^{m-n}\). Put the exponent being subtracted in brackets. \[\frac{x^{-2}}{x^{-5}}=x^{-2-(-5)}=x^{-2+5}=\boxed{x^3}.\] Subtracting \(-5\) means adding \(5\). Do not replace \(-2-(-5)\) by \(-2-5\).

Worked example 6 | Multiply, do not add

Question:  Simplify \((2^3)^{-4}\).

Solution

Use \((a^m)^n=a^{mn}\): the outer power acts on the whole inner power. \[\begin{aligned} (2^3)^{-4}&=2^{3\times(-4)} &&\text{multiply the exponents}\\ &=2^{-12} &&\text{because }3\times(-4)=-12\\ &=\frac1{2^{12}}=\boxed{\frac1{4096}} &&\text{use the reciprocal rule.} \end{aligned}\]

Worked example 7 | Include the numerical coefficient

Question:  Simplify

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

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

Solution

Use \((ab)^n=a^nb^n\) and \((a/b)^n=a^n/b^n\).

(a)  The brackets include both 2 and \(x\), so both receive the power 4: \[(2x)^4=(2x)(2x)(2x)(2x)=2^4x^4=\boxed{16x^4}.\]

(b)  Square the numerator and the denominator, then square each factor in the numerator: \[\left(\frac{3x}{2}\right)^2=\frac{(3x)^2}{2^2}=\frac{3^2x^2}{2^2}=\boxed{\frac{9x^2}{4}}.\]

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

Step 1: expand the numerator. Use \((ab)^n=a^nb^n\) and \((a^m)^n=a^{mn}\): \[(2x^{-3}y^2)^2=2^2(x^{-3})^2(y^2)^2=4x^{-6}y^4.\] Step 2: divide like factors. Use \(a^m/a^n=a^{m-n}\) separately for \(x\) and \(y\): \[\begin{aligned} \frac{4x^{-6}y^4}{4x^{-1}y^{-3}} &=\frac44\times\frac{x^{-6}}{x^{-1}}\times\frac{y^4}{y^{-3}}\\ &=1\times x^{-6-(-1)}\times y^{4-(-3)}\\ &=x^{-6+1}y^{4+3}=x^{-5}y^7. \end{aligned}\] Step 3: use positive exponents. Since \(x^{-5}=1/x^5\), \[x^{-5}y^7=\frac1{x^5}\times y^7=\boxed{\frac{y^7}{x^5}}.\] Only the factor with a negative exponent moves to the denominator. Keep the original restrictions \(x,y\ne0\).

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 \(a^x=b\Longleftrightarrow\log_a b=x\). Identify three things: the base, the exponent, and the result.

(a)  In \(2^5=32\), the base is 2, the exponent is 5, and the result is 32. \[\underbrace{2}_{\text{base}}^{\,5}=32 \quad\Longrightarrow\quad\boxed{\log_2 32=5}.\] Read this as: “The power of 2 needed to produce 32 is 5.”

(b)  In \(\log_{10}0.01=-2\), the base is 10 and the required exponent is \(-2\): \[\boxed{10^{-2}=0.01}.\] Check: \(10^{-2}=1/10^2=1/100=0.01\).

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. \[\begin{aligned} \text{(a)}\quad 10^3=1000&\quad\Rightarrow\quad \log1000=3,\\ \text{(b)}\quad 10^{-3}=0.001&\quad\Rightarrow\quad \log0.001=-3,\\ \text{(c)}\quad 10^0=1&\quad\Rightarrow\quad \log1=0. \end{aligned}\]

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. \[\begin{aligned} \text{(a)}\quad e^1=e&\quad\Rightarrow\quad \ln e=1,\\ \text{(b)}\quad e^0=1&\quad\Rightarrow\quad \ln1=0,\\ \text{(c)}\quad e^4=e^4&\quad\Rightarrow\quad \ln(e^4)=4,\\ \text{(d)}\quad \frac1{e^2}=e^{-2}&\quad\Rightarrow\quad \ln(1/e^2)=-2. \end{aligned}\] 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

Recall \(\ln b=\log_e b\), so \(e^x=b\Longleftrightarrow x=\ln b\).

(a)  The base is \(e\) and the unknown is its exponent. Therefore \[e^x=7\quad\Longrightarrow\quad\log_e7=x\quad\Longrightarrow\quad\boxed{x=\ln7}.\]

(b)  First make the base visible: \(\ln y=-2\) means \(\log_e y=-2\). Thus \[\log_e y=-2\quad\Longrightarrow\quad e^{-2}=y \quad\Longrightarrow\quad\boxed{y=\frac1{e^2}}.\] A negative logarithm is an exponent below zero; it does not mean \(y\) 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.

For \(1.698970\ldots\), the first three significant digits are 1, 6 and 9. The next digit is 8, so rounding gives \(1.70\); keep the final zero to show 3 significant figures.

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

TI-Nspire CX: Evaluate logarithms

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

Step 1: identify the unknown. We need the exponent on 10, so use the conversion \(a^x=b\Longleftrightarrow x=\log_a b\). \[10^x=350\quad\Longrightarrow\quad\boxed{x=\log_{10}350=\log350}\quad\text{(exact)}.\] Step 2: evaluate. Enter log(350) in Scratchpad Calculate: \[x=2.544068044\ldots\quad\Longrightarrow\quad\boxed{x\approx2.54}\quad\text{(3 s.f.)}.\] Step 3: check. Since \(100<350<1000\), we need \(2<x<3\). To check the conversion, enter 10^(log(350)); this returns 350, up to numerical rounding. Using the rounded \(2.54\) instead gives only an approximate check.

TI-Nspire CX: Find and check an exponent

Worked example 15 | Recover the number from its logarithm [GDC]

Question:  Find \(y\), to 3 significant figures, when

1. \(\log y=1.6\);

2. \(\ln y=-0.8\).

Solution

The answer to a logarithm is an exponent. Use that exponent with the correct base.

(a)  \(\log y=\log_{10}y\). Using \(\log_a b=x\Longleftrightarrow b=a^x\), \[\log_{10}y=1.6\quad\Longrightarrow\quad y=10^{1.6}=39.810717\ldots \quad\Longrightarrow\quad\boxed{y\approx39.8}.\] In Scratchpad, enter 10^(1.6). Check that \(\log(39.810717\ldots)\approx1.6\).

(b)  \(\ln y=\log_e y\), so \[\log_e y=-0.8\quad\Longrightarrow\quad y=e^{-0.8}=0.449328\ldots \quad\Longrightarrow\quad\boxed{y\approx0.449}.\] Use the calculator’s \(e^x\) template or insert its built-in constant \(e\) from the symbols palette/Catalog, then raise it to \(-0.8\). Use the negative-number key for the exponent. Do not enter a variable named e or round \(e\) to 2.72. Request a decimal with Ctrl + Enter.

Both values of \(y\) are positive, as required for a real logarithm.

TI-Nspire CX: Undo a logarithm

Worked example 16 | Use a supplied model [GDC]

Question:  A simplified sound model is \(L=10\log R\), where \(L\) is a sound level in decibels (dB) and \(R>0\) is an intensity ratio.

1. Find \(L\) when \(R=400\), to 1 decimal place.

2. Find \(R\) when \(L=20\).

Solution

(a)  Write the model first, substitute, then calculate: \[\begin{aligned} L&=10\log R\\ &=10\log400\\ &=10(2.602059991\ldots)=26.02059991\ldots\\ &\approx\boxed{26.0\text{ dB}}\quad\text{(1 d.p.).} \end{aligned}\] Enter the whole expression 10*log(400). Do not round the logarithm before multiplying.

(b)  Substitute \(L=20\) and divide both sides by 10: \[20=10\log R\quad\Longrightarrow\quad2=\log R \quad\Longrightarrow\quad R=10^2=\boxed{100}.\] The intensity is 100 times the reference intensity. No logarithm law is needed: use the definition.

Formula booklet: this sound model is supplied for the example; it is not a section 1.5 formula to memorise.