Properties of Exponents

The exponent \(2^3\) shows that \(2\) is to multiply itself \(3\) times:

\[2^3=2\times2\times2\]

In \(5^7\), \(7\) is the power or the exponent while \(5\) is the base.

Expand \(2^3\times 2^2\).

Solution:

\[2^3\times 2^2=2^{3+2}=2^5\]

Verification

\[2^3\times 2^2=(2\times 2\times 2)\times (2\times 2)=2^5\]

Expand \(a^4\times a^6\).

Solution:

\[a^4+a^6=a^{4+6}=a^{10}\]

Verification

\[a^4\times a^6=(a\times a\times a\times a)\times (a\times a\times a\times a\times a\times a)=a^{10}\]

Expand \(2^3\times 3^3 \times 2^4\).

Solution:

\[2^3\times 3^3\times 2^4=(2^3\times 2^4)\times 3^3=2^{3+4}\times 3^3=2^7\times 3^3\]

Simplify \)\dfrac{2^3}{2}\).

Solution:

\[\dfrac{2^3}{2}=2^{3-1}=2^2\]

Verification

\[\dfrac{2^3}{2}=\dfrac{2\times 2\times 2}{2}=2\times 2=2^2\]

Simplify \(\dfrac{b^7}{b^3}\).

Solution:

\[\dfrac{b^7}{b^3}=b^{7-3}=b^4\]

Verification

\[\dfrac{b^7}{b^3}=\dfrac{b\times b\times b\times b\times b\times b\times b}{b\times b\times b}=b\times b\times b\times b=b^4\]

Simplify \(\dfrac{2^5\times 3^7}{3^4}\).

Solution:

\[\dfrac{2^5\times 3^7}{3^7}=2^5\times \dfrac{3^7}{3^4}=2^5\times 3^{7-4}=2^5\times 3^3\]

a. \(2^0\)

b. \(-2^0\)

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

Solution:

a. Anything to the power of \(0\) is equal to \(1\). So we have

\[2^0=1\]

b. Here, the base is \(2\), with a \(-1\) multiplied out front. It becomes

\[\begin{align}-2^0&=-1\times 2^0=\\&=-1\times 1=\\&=-1\end{align}\]

c. Here, the base is \(-2\), and anything to the power of \(0\) equals \(1\). It becomes

\[(-2)^0=1\]

\[2^{-1}=\dfrac{1}{2}\]

\[9^{-3}=\left(\dfrac{1}{9}\right)^3\]

\[6^{-5}=\left(\dfrac{1}{6}\right)^5\]

Find the value of the following expressions.

a. \(125^{\frac{1}{3}}\)

b. \(2^{\frac{4}{3}}\)

c. \(16^{-\frac{3}{2}}\)

Solution:

a. The first step is to express \(125\) as a product of its prime factors,

\[125=5\times 5\times 5=5^3\]

Thus we have,

\[125^{\frac{1}{3}}=\sqrt[3]{125}=\sqrt[3]{5^3}=5\]

b.

\[2^{\frac{4}{3}}=\sqrt[3]{2^4}=\sqrt[3]{16}\]

c. There are two ways to solve this question and both ways involve using the fractional exponent rule and the negative exponent rule.

First, use the fractional exponent rule.

\[\begin{align}16^{-\frac{3}{2}}&=\left(\sqrt[2]{16}\right)^{-3}=\\&=(4)^{-3}\end{align}\]

From here, we use the negative exponent rule

\[\begin{align}(4)^{-3}&=\dfrac{1}{4^3}=\\&=\dfrac{1}{64}\end{align}\]

Solving it in an alternative way, you will first use the negative exponent rule.

\[16^{-\frac{3}{2}}=\dfrac{1}{16^{\frac{3}{2}}}\]

We will now use the fractional exponent rule on the denominator.

\[\begin{align}\dfrac{1}{16^{\frac{3}{2}}}&=\dfrac{1}{\left(\sqrt[2]{16}\right)^3}=\\&=\dfrac{1}{4^3}=\\&=\dfrac{1}{64}\end{align}\]

You get the same answer!

Verify that \(6^3=2^3\times 3^3\).

Solution:

Method 1

On the one hand, we have; \(6^3=6\times 6\times 6=216\).

On the other hand, \(2^3\times 3^3=8\times 27=216\).

Thus \(6^3=2^3\times 3^3\).

Method 2

\[6^3=(2\times 3)^2=2^3\times 3^3\]

Verify that \(2^3=\dfrac{6^3}{3^3}\).

Solution:

Method 1

To begin,

\[2^3=2\times 2\times 2=8\]

Also,

\[\begin{align}\dfrac{6^3}{3^3}&=\dfrac{6\times 6\times 6}{3\times 3\times 3}=\\\\&=\dfrac{^2\cancel{6}\times ^2\cancel{6}\times ^2\cancel{6}}{^1\cancel{3}\times ^1\cancel{3}\times ^1\cancel{3}}=\\\\&= 2\times 2\times 2=\\\\&=8\end{align}\]

This implies that

\[2^3=\dfrac{6^3}{3^3}\]

Method 2

Apply the quotient property;

\[\begin{align}\dfrac{6^3}{3^3}&=\left(\dfrac{6}{3}\right)^3=\\&=2^3=\\&=8\end{align}\]

Therefore;

\[2^3=\dfrac{6^3}{3^3}\]

Verify that \(\left(2^3\right)^2=2^6\).

Solution:

Method 1

On the one hand, we have

\[2^3=2\times 2\times 2=8\]

Thus

\[\left(2^3\right)^2=8^2=64\]

On the other hand,

\[2^6=2^{3+3}=2^3\times 2^3=8\times 8=64\]

Thus,

\[\left(2^3\right)^2=2^6=64\]

Method 2

\[\left(2^3\right)^2=2^{3\times 2}=2^6=64\]

Calculate the following without the use of calculators.

a. \((-3x^3y^2)(2x^6y^5)\)

b. \((2b)^{-4}\)

d. \(81^{\frac{3}{4}}\)

e. \(\dfrac{-12m^4n^3(m^3n^2)}{36m^7n^5}\)

Solution:

a. For the expression,

\[(-3x^3y^2)(2x^6y^5)\]

We express them as separate products,

\[(-3x^3y^2)(2x^6y^5)=(-3\times x^3\times y^2)\times (2\times x^6\times y^5)\]

We expand the brackets,

\[(-3x^3y^2)(2x^6y^5)=-3\times x^3\times y^2\times 2\times x^6\times y^5\]

Next, we bring like terms together,

\[\begin{align}(-3x^3y^2)(2x^6y^5)&=-3\times 2\times x^3\times x^6\times y^2\times y^5=\\&=-6\times \left(x^{3+6}\right)\times\left(y^{2+5}\right)=\\&=-6\times x^9\times y^7=\\&=-6x^9y^7\end{align}\]

b. For the expression,

\[(2b)^{-4}\]

We first get rid of the negative exponent, we apply the reciprocal rule,

\[(2b)^{-4}=\dfrac{1}{(2b)^4}=\dfrac{1}{2^4b^4}=\dfrac{1}{16b^4}\]c. For the expression,

\[\left(\dfrac{-6x^6}{3x^3}\right)^{-2}\]

To get rid of the negative exponent, we apply the reciprocal rule,

\[\left(\dfrac{-6x^6}{3x^3}\right)^{-2}=\left(\dfrac{3x^3}{-6x^6}\right)^2\]

We then divide the like terms of the expression in the bracket,

\[\left(\dfrac{-6x^6}{3x^3}\right)^{-2}=\left(\dfrac{3x^3}{-6x^6}\right)^2=\left(-\dfrac{1}{2}\left(x^{3-6}\right)\right)^2=\left(-\dfrac{1}{2}x^{-3}\right)^2\]

Afterward, we distribute the exponent \(2\) to the product inside the bracket to get,

\[\begin{align}\left(\dfrac{-6x^6}{3x^3}\right)^{-2}=\left(-\dfrac{1}{2}\times \dfrac{1}{x^3}\right)^2=\\&=\left(-\dfrac{1}{2x^3}\right)^2=\\&=\dfrac{(-1)^2}{(2x^3)^2}=\\&=\dfrac{1}{2^2(x^3)^2}=\\&=\dfrac{1}{4x^6}\end{align}\]

d. For the expression,

\[81^{\frac{3}{4}}\]

we recall first the fraction exponent rule,

\[81^{\frac{3}{4}}=\left(\sqrt[4]{81}\right)^3=\sqrt[4]{81^3}\]

But

\[81=9^2=\left(3^2\right)^2=3^4\]

Hence

\[81^{\frac{3}{4}}=\sqrt[4]{81^3}=\sqrt[4]{\left(3^4\right)^3}=\sqrt[4]{\left(3^3\right)^4}=3^3\]

We can see in another way, we recall that,

\[\left(a^m\right)=a^{mn}\]

Thus,

\[81^{\frac{3}{4}}=\left(3^4\right)^{\frac{3}{4}}=3^{4\times \frac{3}{4}}=3^3\]

e. For the expression,

\[\dfrac{-12m^4n^3(m^3n^2)}{36m^7n^5}\]

We first expand the numerator,

\[\dfrac{-12m^4n^3(m^3n^2)}{36m^7n^5}=\dfrac{-12\times m^4\times n^3\times m^3\times n^2}{36m^7n^5}\]

We next bring like terms together, to get

\[\begin{align}\dfrac{-12m^4n^3(m^3n^2)}{36m^7n^5}&=\dfrac{-12\times m^4\times m^3\times n^3\times n^2}{36m^7n^5}=\\&=\dfrac{-12\times m^{4+3}\times n^{3+2}}{36m^7n^5}=\\&=\dfrac{-12\times m^7\times n^5}{36m^7n^5}=\\&=\dfrac{-12\times m^7\times n^5}{36\times m^7\times n^5}\end{align}\]

We then divide like terms to get,

\[\begin{align}\dfrac{-12m^4n^3(m^3n^2)}{36m^7n^5}&=\left(\dfrac{-12}{36}\right)\times \left(\dfrac{m^7}{m^7}\right)\times\left(\dfrac{n^5}{n^5}\right)=\\&=\left(-\dfrac{1}{3}\right)\times m^{7-7}\times n^{5-5}=\\&=\left(-\dfrac{1}{3}\right)\times m^0\times n^0\end{align}\]

\[\dfrac{-12m^4n^3(m^3n^2)}{36m^7n^5}=\left(-\dfrac{1}{3}\right)\times 1\times 1=-\dfrac{1}{3}\]

Simplify the expression \(\dfrac{m^{\frac{3}{4}}\times n^{\frac{1}{2}}}{m^{\frac{1}{2}}\times n^{-\frac{3}{2}}}\).

Solve when \(m= 16\) and \(n= 3\).

Solution:

\[\begin{align}\dfrac{m^{\frac{3}{4}}\times n^{\frac{1}{2}}}{m^{\frac{1}{2}}\times n^{-\frac{3}{2}}}&=\dfrac{m^{\frac{3}{4}}}{m^{\frac{1}{2}}}\times \dfrac{n^{\frac{1}{2}}}{n^{-\frac{3}{2}}}=\\&=\left(m^{\frac{3}{4}}\div m^{\frac{1}{2}}\right)\times \left(n^{\frac{1}{2}}\div n^{-\frac{3}{2}}\right)=\\&=m^{\frac{3}{4}-\frac{1}{2}}\times n^{\frac{1}{2}-\left(-\frac{3}{2}\right)}=\\&=m^{\frac{1}{4}}\times n^2\end{align}\]

Substitute the value of \(m\) as \(16\) and \(n\) as \(3\) into the expression;

\[\begin{align}m^{\frac{1}{4}}\times n^2&=16^{\frac{1}{4}}\times 3^2=\\&=\left(2^4\right)^{\frac{1}{4}}\times 3^2=\\&=2^{4\times \frac{1}{4}}\times 3^2=\\&=2\times 9=\\&=18\end{align}\]

Convert \(38 000 000 000\) meters per second to Scientific Notation.

Solution:

The first step is to count from your left to right. We have the number \(38 000 000 000\).

We have \(38\) and \(9\) zeros to the right of it.

We recall the scientific notation,

\[q\times 10^p\]

\[1\leq q<10\]

where \(p\) is an integer. Thus

\[38 000 000 000=38\times 10^9\]

Now \(38\) should be written as \(q\times 10^p\) as well. Thus,

\[38=3.8\times 10\]

Now we replace in the initial expression to get,

\[38 000 000 000=3.8\times 10\times 10^9=3.8\times 10^{10}\]

The length and bread of a rectangular mark is \(2\, \text{mm}\) and \(6\, \text{mm}\) respectively, calculate the perimeter in kilometers leaving your answer in standard form.

Solution:

The perimeter of a rectangle is given as

\[\begin{align}\text{Perimeter of the mark}&=2\times\left(\text{length}+\text{breadth}\right)=\\&=2\times\left(2\,\text{mm}+6\,\text{mm}\right)=\\&=2\times 8\,\text{mm}=\\&=16\,\text{mm}\end{align}\]

Recall that;

\[1000\,\text{m}=1\text{ km}\]

\[100\text{ cm}=1\text{ m}\]

\[10\text{ mm}=1\text{ cm}\]

\[100\times 10\text{ mm}=1\text{ m}\]

\[1000\times 100\times 10 \text{ mm}=1\text{ km}\]

\[1\times 10^6\text{ mm}=1\text{ km}\]

\[\dfrac{1\times 10^6\text{ mm}}{10^6}=\dfrac{1\text{ km}}{10^6}\]

\[\dfrac{1\times \cancel{10^6}\text{ mm}}{\cancel{10^6}}=\dfrac{1\text{ km}}{10^6}\]

\[1\text{ mm}=\dfrac{1}{10^6}\text{ km}\]

Recall that;

\[a^{-1}=\dfrac{1}{a}\]

Thus;

\[\dfrac{1}{10^6}=10^{-6}\]

This means that;

\[1\text{ mm}=10^{-6}\text{ km}\]

So convert \(16\text{ mm}\) to \(\text{km}\);

\[1\text{ mm}=10^{-6}\text{ km}\]

\[1\text{ mm}\times 16=10^{-6}\text{ km}\times 16\]

\[16\text{ mm}=16\times 10^{-6}\text{ km}\]

Now \(16\) should be written as \(q\times 10^p\) as well. Thus,

\[16=1.6\times 10\]

Now we replace it in the initial expression to get;

\[\begin{align}16\times 10^{-6}\text{ km}&=1.6\times 10\times 10^{-6}\text{ km}=\\&=1.6\times 10^1\times 10^{-6}\text{ km}=\\&=1.6\times 10^{-6+1}\text{ km}=\\&=1.6\times 10^{-5}\text{ km}\end{align}\]

\[\text{Perimeter of the mark}=1.6\times 10^{-5}\text{ km}\]