Math Practice Sheets for Exponents and Logarithms

Creation Date: October 24, 2024

Contributors

Erica Madebeykin

PGDip Researcher

Viktar Satanchuk

Data Scientist

1. Simplifying Equations with Exponents

NOTE: Ongoing Development

1.1 Using Roots to solve Exponents

Equation 1: Solve for x

$$x^2=4$$

Click for Solution:

Taking the square root of both sides:

$$\sqrt{x^2}=\sqrt{4}$$

The square root cancels the exponent of 2 on the left side, while also taking the square root of 4 on the right side:

$$\cancel{\sqrt{}}{x}^{\cancel{2}}=\sqrt{4}$$

Thus, we conclude the solutions are: x = 2 or x = -2

$$x=\pm 2$$

1.2 Using the Multiplicative Inverse Formula

Solve the equation #1:

$$15\cdot {({0.5}^{-1}+{0.8}^{-1})}^{-1}\cdot (2^{-1}+{13}^{-1}{)}^{-1}$$

You can use the Multiplicative Inverse Formula:

$$x^{-1}\cdot x=1$$

Very Long Solution

To solve this, we can break it down into manageable parts. Think of the first part is 15, while the others represent the second and third parts of our expression.

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Working on the Second Part

Now let's start by solving the second part:

$$({0.5}^{-1}+{0.8}^{-1}{)}^{-1}$$

To simplify this, we’ll evaluate each term individually using the Multiplicative Inverse formula. Multiplicative Inverse formula:

$$x^{-1}\cdot x=1$$

Step 1: Simplifying 0.5 raised to negative 1 We start with:

$${0.5}^{-1}\to {0.5}^{-1}\cdot x=1$$

We can solve this by using the Multiplicative Inverse Formula

Now Solving for x:

$$x=\frac{1}{0.5}=\frac{1}{\frac{5}{10}}=\frac{1}{\frac{1}{2}}=\frac{1}{1}\cdot \frac{2}{1}=2$$

Now we solve for 0.8 raised to negative 1 by using again the same multiplicative inverse formula

$$\begin{array}{rl}{0.8}^{-1}& \to {0.8}^{-1}\cdot x=1\\ & \to x=\frac{1}{0.8}=\frac{1}{\frac{8}{10}}=\frac{1}{\frac{4}{5}}\\ & =\frac{1}{1}\cdot \frac{5}{4}=\frac{5}{4}=1\frac{1}{4}\text{ or }1.25\end{array}$$

Hence adding these two pre-simplification for the 2nd part equation:

$$(2+1.25{)}^{-1}=(3.25{)}^{-1}$$

Now we have to further simplfy by:

$$\begin{array}{rl}{3.25}^{-1}& \to {3.25}^{-1}\cdot x=1\\ & \to x=\frac{1}{3.25}=\frac{1}{3\frac{1}{4}}=\frac{1}{\frac{13}{4}}\\ & =\frac{4}{13}\end{array}$$

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To incorporate these individual simplifcation as the 2nd part: $$=15\cdot \frac{4}{13}\cdot (2^{-1}+{13}^{-1}{)}^{-1}$$

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Working on the Third Part

Now we start working at the 3rd part of the equation that is

$$(2^{-1}+{13}^{-1}{)}^{-1}$$

Simplifying the operation inside the parenthesis:

$$(2^{-1})\to 2^{-1}\cdot x=1\to x=\frac{1}{2}$$$$({13}^{-1})\to {13}^{-1}\cdot x=1\to x=\frac{1}{13}$$

Now we have:

$$(\frac{1}{2}+\frac{1}{13}{)}^{-1}$$

To add these 2 fractions, we find the common denominator which is 26 (because 2 x 13 is 26). So we do:

$$\frac{26}{13}=2;\frac{26}{2}=13$$

So now we can translate and add them like this:

$$\begin{array}{rl}{(\frac{1}{2}+\frac{1}{13})}^{-1}& \to {(\frac{13}{26}+\frac{2}{26})}^{-1}\\ & ={\left(\frac{15}{26}\right)}^{-1}=\frac{1}{\frac{15}{26}}\\ & =\frac{1}{1}\cdot \frac{26}{15}=\frac{26}{15}\end{array}$$

Therefore, our 3rd part of the equation results to 26/15

Multiplication of all 3 parts of the equation

$$15\cdot \frac{4}{13}\cdot \frac{26}{15}$$

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Multiply 15 by 4/13

$$\frac{15}{1}\cdot \frac{4}{13}=\frac{15\cdot 4}{13}=\frac{60}{13}$$

Now multiply the product by the next fraction in line, the 3rd part which is 26/15

$$\frac{60}{13}\cdot \frac{26}{15}=\text{?}$$

We Simplify

15 and 60 share the common factor of 15 because 15 x 4 = 60.

$$\frac{60}{15}=4\frac{15}{15}=1$$

So we can cancel them by computing them like this:

$$\frac{{\cancel{60}}^4}{13}\cdot \frac{26}{{\cancel{15}}^1}=\frac{4}{13}\cdot \frac{26}{1}$$

thus, we have:

$$=\frac{104}{13}$$$$=8$$

So our final answer is 8.

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Equation #2

$$9\cdot ({0.3}^{-1}+{0.6}^{-1}{)}^{-1}\cdot (4^{-1}+5^{-1}{)}^{-1}$$

Long Solution

We divide it into 3 parts: First Part is 9
Second Part is

$$\cdot ({0.3}^{-1}+{0.6}^{-1}{)}^{-1}$$

Third Part is

$$\cdot (4^{-1}+5^{-1}{)}^{-1}$$

Working on the 2nd Part

$$({0.3}^{-1}+{0.6}^{-1}{)}^{-1}$$$$\begin{array}{rl}{0.3}^{-1}\to \frac{1}{0.3}\cdot x=1;& \text{given that 0.3 means}\frac{3}{10}\\ & \text{so}x=\frac{1}{\frac{3}{10}}\\ & =\frac{10}{3}\end{array}$$$$\begin{array}{rl}{0.6}^{-1}\to \frac{1}{0.6}\cdot x=1;& \text{given that 0.6 means}\frac{6}{10}\\ & \text{so}x=\frac{1}{\frac{6}{10}}\\ & =\frac{10}{6}\text{or}\frac{5}{3}\end{array}$$

Combining these two for the 2nd part:

$$(\frac{10}{3}+\frac{5}{3}{)}^{-1}=(\frac{15}{3}{)}^{-1}=5^{-1}=\frac{1}{5}$$

Working on the 3rd Part

$$(4^{-1}+5^{-1}{)}^{-1}$$$$4^{-1}\to \frac{1}{4}$$$$5^{-1}\to \frac{1}{5}$$

So

$$(\frac{1}{4}+\frac{1}{5}{)}^{-1}\to (\frac{5}{20}+\frac{4}{20}{)}^{-1}=(\frac{9}{20}{)}^{-1}$$

To simplify this, we use again the multiplicative inverse strategy:

$$x=\frac{1}{\frac{9}{20}}=\frac{20}{9}\cdot \frac{1}{1}=\frac{20}{9}$$

Combining all the three parts

$$9\cdot \frac{1}{5}\cdot \frac{20}{9}=?$$

First part times the Second Part:

$$\frac{9}{\cancel{1}}\cdot \frac{1}{5}=\frac{9}{5}$$

then we multiplied it to the 3rd part:

$$\frac{9}{5}\cdot \frac{20}{9}=\frac{{\cancel{9}}^1}{{\cancel{5}}^1}\cdot \frac{{\cancel{20}}^4}{{\cancel{9}}^1}=\frac{4}{1}=4$$

So our final answer is 4 .

1.3 More practice on negative exponents

Solve the Equation #1:

$$(-{0.5}^{-2}+{0.2}^{-2})(3^{-1}+7^{-1})$$

Solution

We solve this again, part by part.
Part 1 is

$$(-{0.5}^{-2}+{0.2}^{-2})$$

Part 2 is

$$(3^{-1}+7^{-1})$$

Solving the first part

$$\begin{array}{rl}(-{0.5}^{-2}+{0.2}^{-2})& \to \frac{1}{{0.5}^2}+\frac{1}{{0.2}^2}\\ & =-\frac{1}{0.25}+\frac{1}{0.04}\\ & =-4+25\\ & =21\end{array}$$

Solving the Second part which is easier

$$\begin{array}{rl}(3^{-1}+7^{-1})& \to \frac{1}{3}+\frac{1}{7}\\ & =\frac{7}{21}+\frac{3}{21}=\frac{10}{21}\end{array}$$

Combining the First Part and Second Part together

$$21\cdot \frac{10}{21}=\frac{\cancel{21}}{1}\cdot \frac{10}{\cancel{21}}=\frac{10}{1}=10$$

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Solve the Equation #2

$$((-1\frac{1}{3}{)}^2+(4\frac{1}{2}{)}^{-1})\cdot {1.5}^{-2}\cdot (\frac{1}{3}{)}^{-2}$$