Functions and algebra: Solve algebraic equations and inequalities

# Unit 2: Solve exponential and literal equations

Dylan Busa ### Unit outcomes

By the end of this unit you will be able to:

• Solve exponential equations of the form $\scriptsize k{{a}^{{x+p}}}=m$ by using the laws of exponents.
• Solve literal equations.

## What you should know

Before you start this unit, make sure you can:

• Simplify expressions containing exponents using the exponent laws. If you need to revise the exponent laws see Subject outcome 1.2, Units 2 and 3.
• Solve linear equations. If you need to revise solving linear equations, see Unit 1 in this subject outcome.

Here is a short self-assessment to make sure you have the skills you need to proceed with this unit.

1. Simplify the following:
1. $\scriptsize {{2}^{x}}{{.2}^{3}}$
2. $\scriptsize {{\left( {{{3}^{3}}} \right)}^{{3x+4}}}$
3. $\scriptsize {{4}^{{2x-3}}}\times {{4}^{{3x+1}}}$
2. Solve for $\scriptsize x$:
1. $\scriptsize \displaystyle \frac{4}{x}+12=4$
2. $\scriptsize \displaystyle \frac{{3x}}{2}+\displaystyle \frac{{5x}}{3}=5$

Solutions

1. .
1. $\scriptsize {{2}^{x}}{{.2}^{3}}={{2}^{{x+3}}}$
2. $\scriptsize {{\left( {{{3}^{3}}} \right)}^{{3x+4}}}={{3}^{{9x+12}}}$
3. $\scriptsize {{4}^{{2x-3}}}\times {{4}^{{3x+1}}}={{4}^{{(2x-3)+(3x+1)}}}={{4}^{{5x-2}}}$
2. Solve for $\scriptsize x$:
1. .
\scriptsize \begin{align*}\displaystyle \frac{4}{x}+12 & =4\\\therefore \displaystyle \frac{4}{x} & =-8\\\therefore -8x & =4\\\therefore x & =-\displaystyle \frac{1}{2}\end{align*}
2. .
\scriptsize \begin{align*}\displaystyle \frac{{3x}}{2}+\displaystyle \frac{{5x}}{3} & =5\quad \text{Multiply both sides by the LCD of }6\\\therefore 9x+10x & =30\\\therefore 19x & =30\\\therefore x & =\displaystyle \frac{{30}}{{19}}\end{align*}

## Introduction

At the start of the previous unit, we said that equations are at the heart of almost all Mathematics. In that unit we learnt how to solve linear and quadratic equations. Linear equations are equations where the exponent on the unknown is $\scriptsize 1$. Linear equations only ever have one solution. An example of a linear equation is $\scriptsize 4x+5=17$. The exponent on the unknown $\scriptsize x$ is $\scriptsize 1$.

We also learnt how to solve quadratic equations. These are equations where the highest power on the unknown is $\scriptsize 2$ and they have up to two solutions. An example of a quadratic equation is $\scriptsize {{x}^{2}}+4x=-3$. Here the highest exponent on the $\scriptsize x$ is $\scriptsize 2$.

In principle, we can also solve equations where the highest exponent on the unknown is $\scriptsize 3$ (cubic equations), $\scriptsize 4$, $\scriptsize 5$ or indeed any natural number $\scriptsize n$. Together, these are called polynomial equations. However, in all these cases, the unknown is in the base not the exponent.

In this unit, we are going to solve a different kind of equation, where the unknown is in the exponent. These are called exponential equations. ### Example 2.1

Solve for $\scriptsize x$: $\scriptsize {{3}^{x}}-27=0$.

Solution:

$\scriptsize {{3}^{x}}-27=0$

As usual, the first step in solving an equation is to isolate the unknown. In this case, we have to isolate the term with the unknown in it, $\scriptsize {{3}^{x}}$. We add $\scriptsize 27$ to both sides of the equation.

$\scriptsize \therefore {{3}^{x}}=27$

Now you might already be able to see what the solution is. The general principle when solving exponential equations is to get the bases on both sides of the equation the same. We know that $\scriptsize 27={{3}^{3}}$.

$\scriptsize \therefore {{3}^{x}}={{3}^{3}}$

Now we can equate the exponents on both sides of the equation to solve for the unknown.
$\scriptsize \therefore x=3$

Here are three more examples for you to work through. ### Example 2.2

Solve for $\scriptsize x$: $\scriptsize {{3}^{{x-2}}}-45=36$.

Solution:

$\scriptsize {{3}^{{x-2}}}-45=36$

Once again, we need to isolate the terms with the unknown by adding $\scriptsize 45$ to both sides.
\scriptsize \begin{align*}\therefore {{3}^{{x-2}}}=36+45\\\therefore {{3}^{{x-2}}}=81\end{align*}

Next, we need to get the bases on both sides the same. We know that $\scriptsize 81={{3}^{4}}$.
$\scriptsize \therefore {{3}^{{x-2}}}={{3}^{4}}$

Now we can equate the exponents and solve for the unknown.

\scriptsize \begin{align*}\therefore x-2=4\\\therefore x=6\end{align*} ### Example 2.3

Solve for $\scriptsize y$: $\scriptsize {{5}^{y}}+{{3.5}^{{y+1}}}=400$.

Solution:

$\scriptsize {{5}^{y}}+{{3.5}^{{y+1}}}=400$

We have two terms with unknowns in the exponent that we need to combine. To do this, we need to recognise that $\scriptsize {{5}^{{y+1}}}={{5}^{y}}{{.5}^{1}}$. Now we can simplify the second term on the left-hand side.

\scriptsize \begin{align*}\therefore {{5}^{y}}+{{3.5.5}^{y}}=400\\\therefore {{5}^{y}}+{{15.5}^{y}}=400\end{align*}

On the left-hand side we have two terms that we can add. If you need to, you can substitute $\scriptsize {{5}^{y}}$ for $\scriptsize \text{A}$ if this makes the step easier.

$\scriptsize \therefore {{16.5}^{y}}=400$ or
\scriptsize \begin{align*}\therefore {{5}^{y}}+{{15.5}^{y}}&=400\quad \text{Let }{{5}^{y}}\text{ be A}\\\therefore \text{A}+15\text{A}&=400\\\therefore 16\text{A}&=400\quad \text{Substitute back for }{{5}^{y}}\\\therefore {{16.5}^{y}}&=400\end{align*}

Now we can isolate the term with the unknown by dividing both sides by $\scriptsize 16$.

$\scriptsize \therefore {{5}^{y}}=25$

From here on, it is easy to solve for $\scriptsize y$.

\scriptsize \begin{align*}\therefore {{5}^{y}}={{5}^{2}}\\\therefore y=2\end{align*}

### Note

$\scriptsize {{3}^{x}}+{{3}^{{2x}}}\ne {{2.3}^{x}}$ and $\scriptsize {{3}^{x}}+{{3}^{{2x}}}\ne {{2.3}^{{2x}}}$
$\scriptsize {{3}^{x}}\times {{3}^{{2x}}}={{3}^{{x+2x}}}={{3}^{{3x}}}$ ### Example 2.4

Solve for $\scriptsize t$: $\scriptsize {{7}^{t}}=50-{{7}^{{t+2}}}$

Solution:

\scriptsize \begin{align*}{{7}^{t}}=50-{{7}^{{t+2}}}\\\therefore {{7}^{t}}+{{7}^{{t+2}}} & =50\quad \text{Get all terms with unknowns in the exponent on one side}\\\therefore {{7}^{t}}+{{7}^{t}}{{.7}^{2}} & =50\quad \text{Split the two term exponent into its components}\\\therefore {{7}^{t}}+{{49.7}^{t}} & =50\quad \text{Collect the like terms}\\\therefore {{50.7}^{t}} & =50\\\therefore {{7}^{t}} & =1\quad \text{Remember that }1={{a}^{0}}\\\therefore {{7}^{t}} & ={{7}^{0}}\\\therefore t & =0\end{align*}

### Note

If $\scriptsize {{a}^{x}}={{a}^{y}}$ then $\scriptsize x=y$ if $\scriptsize a \gt 0$ and $\scriptsize a\ne 1$. ### Exercise 2.1

Solve for the unknown in the following:

1. $\scriptsize {{2}^{{a+5}}}=32$
2. $\scriptsize {{16}^{{2b+5}}}={{64}^{{b+3}}}$
3. $\scriptsize {{81}^{{x+3}}}={{27}^{{x-4}}}$
4. $\scriptsize {{81}^{y}}+{{9}^{{2y+1}}}=270$
5. $\scriptsize {{2}^{x}}+{{2}^{{x+2}}}=40$

The full solutions are at the end of the unit.

## Solve literal equations

Very often in the real world, equations will have more than one variable. Take the equation $\scriptsize E=m{{c}^{2}}$. Although the implications of this equation are hugely significant, the equation itself is very simple. It has the two variables and one constant. The constant is $\scriptsize c$, which is the speed of light in a vacuum, while the variables are $\scriptsize E$ (energy) and $\scriptsize m$ (mass).

We can use this equation as it is to calculate the total amount of energy contained in a given mass. Alternatively, we can rearrange it to calculate the amount of mass needed to produce a given energy.

$\scriptsize m=\displaystyle \frac{E}{{{{c}^{2}}}}$

We call equations like this literal equations. Other common examples are the area of a circle ($\scriptsize A=\pi {{r}^{2}}$) or the volume of a rectangular prism ($\scriptsize V=l\times b\times h$).

When we ‘solve’ literal equations, all we are really doing is isolating one of the variables on one side of the equation. ### Example 2.5

The area of a triangle is $\scriptsize A=\displaystyle \frac{1}{2}\times b\times h$. What is the base of the triangle in terms of its area and height?

Solution:

$\scriptsize A=\displaystyle \frac{1}{2}\times b\times h$

We need to isolate the variable for the base ($\scriptsize b$) on one side of the equation. To do this, we can start by multiplying both sides of the equation by $\scriptsize 2$.

$\scriptsize \therefore 2A=b\times h$

Now we can divide both sides of the equation by $\scriptsize h$.

$\scriptsize \therefore \displaystyle \frac{{2A}}{h}=b$

We normally write the equation with the isolated variable on the left-hand side.

$\scriptsize \therefore b=\displaystyle \frac{{2A}}{h}$ ### Example 2.6

Given the formula $\scriptsize b=G\times \displaystyle \frac{q}{{G+{{r}^{2}}}}$

1. Make $\scriptsize G$ the subject of the formula.
2. Express $\scriptsize r$ in terms of $\scriptsize b$, $\scriptsize G$ and $\scriptsize q$.

Solutions:

1. Make $\scriptsize G$ the subject of the formula is just another way of saying isolate the variable $\scriptsize G$ on one side of the equation.
.
\scriptsize \begin{align*}b & =G\times \displaystyle \frac{q}{{G+{{r}^{2}}}}\\\therefore b & =\displaystyle \frac{{Gq}}{{G+{{r}^{2}}}}\quad \text{Multiply both sides by }G+{{r}^{2}}\\\therefore b(G+{{r}^{2}}) & =Gq\quad \text{Divide both sides by }b\text{ and by }G\\\therefore \displaystyle \frac{{G+{{r}^{2}}}}{G} & =\displaystyle \frac{q}{b}\quad \text{Write the fraction on the left-hand side as two fractions}\\\therefore \displaystyle \frac{G}{G}+\displaystyle \frac{{{{r}^{2}}}}{G} & =\displaystyle \frac{q}{b}\\\therefore 1+\displaystyle \frac{{{{r}^{2}}}}{G} & =\displaystyle \frac{q}{b}\\\therefore \displaystyle \frac{{{{r}^{2}}}}{G} & =\displaystyle \frac{q}{b}-1\quad \text{Divide both sides by }{{r}^{2}}\\\therefore \displaystyle \frac{1}{G} & =\displaystyle \frac{\frac{q}{b}-1}}{{{{r}^{2}}}}\end{align*
.
We need to have $\scriptsize G$ on the left-hand side. Remember that you can do anything to an equation so long as you do the same thing to both sides of the equation. Take the inverse of both sides of the equation which will flip both sides around; the numerator will become the denominator and the denominator will become the numerator.
.
$\scriptsize \therefore G=\displaystyle \frac{{{{r}^{2}}}}{\frac{q}{b}-1}$
2. Being asked to express one variable in terms of others is another way of asking you to isolate that variable on one side of the equation, or to solve the equation for that variable. In Question 1 above, we arrived at the following:
.
\scriptsize \begin{align*}\displaystyle \frac{{{{r}^{2}}}}{G} & =\displaystyle \frac{q}{b}-1\quad \text{Multipy both side by }G\\\therefore {{r}^{2}} & =G\left( \frac{q}{b}-1} \right)\end{align*
.
Now we have $\scriptsize {{r}^{2}}$ on the left-hand side. To get just $\scriptsize r$, we need to take the square root of both sides. Remember when you take the square root, the answer could be positive or negative.
.
$\scriptsize \therefore r=\pm \sqrt{{G\left( \frac{q}{b}-1} \right)}}\quad \text{Remember to take the square root of the whole right-hand side$ ### Exercise 2.2

1. Solve for $\scriptsize y$ in the following: $\scriptsize 2x+3y=9$.
2. Make $\scriptsize a$ the subject of the formula: $\scriptsize s=ut+\displaystyle \frac{1}{2}a{{t}^{2}}$.
3. Solve for $\scriptsize r$: $\scriptsize A=\pi {{R}^{2}}-\pi {{r}^{2}}$.
4. If $\scriptsize r=\sqrt{{{{a}^{2}}+{{b}^{2}}}}$, solve for $\scriptsize b$.
5. $\scriptsize {{F}_{R}}=\displaystyle \frac{1}{2}\rho A{{C}_{D}}{{v}^{2}}$. If$\scriptsize {{F}_{R}}=24$, $\scriptsize A=0.5$, $\scriptsize {{C}_{D}}=0.01$ and $\scriptsize \rho =1.2$, solve for $\scriptsize v$.

The full solutions are at the end of the unit.

### Note

When solving literal equations remember the following:

• To isolate the required variable, perform the operation that will ‘undo’ the operations currently being performed on the variable.
• If the unknown variable is in two or more terms, collect these terms on one side of the equation and try and take it out as a common factor.
• If we have to take the square root of both sides, remember that there will be a positive and a negative answer.
• If the variable that needs to be isolated is in the denominator, start by multiplying both sides by the lowest common denominator.

## Summary

In this unit you have learnt the following:

• How to solve equations where the unknown is in the exponent (exponential equations) by getting the bases on both sides of the equation to be the same using the exponent laws.
• How to rearrange equations with more than one unknown in terms of any of the unknowns.

# Unit 2: Assessment

#### Suggested time to complete: 30 minutes

1. Solve for the unknown in the following:
1. $\scriptsize {{12}^{{2x-3}}}=1$
2. $\scriptsize {{25}^{{1-2x}}}-{{5}^{4}}=0$
3. $\scriptsize -\displaystyle \frac{1}{2}{{.6}^{\frac{m}{2}+3}}}=-1$
2. When measuring the diameter of a large pipe, the formula is $\scriptsize D=L+\displaystyle \frac{{{{W}^{2}}}}{{2L}}$. Solve for $\scriptsize W$.
3. The formula for the volume of a geometric shape is given as $\scriptsize V=\displaystyle \frac{1}{3}\pi {{r}^{2}}h$
1. Make $\scriptsize r$ the subject of the formula.
2. Determine the value of $\scriptsize r$ to two decimal places if $\scriptsize V=250\ \text{m}\text{m}^\text{3}$ and $\scriptsize h=25\ \text{mm}$.

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

1. .
\scriptsize \begin{align*}{{2}^{{a+5}}} & =32\\\therefore {{2}^{{a+5}}} & ={{2}^{5}}\\\therefore a+5 & =5\\\therefore a & =0\end{align*}
2. .
\scriptsize \begin{align*}{{16}^{{2b+5}}} &= {{64}^{{b+3}}}\\\therefore {{\left( {{{4}^{2}}} \right)}^{{2b+5}}} & ={{\left( {{{4}^{3}}} \right)}^{{b+3}}}\quad \text{Remember that }{{\left( {{{a}^{m}}} \right)}^{n}}={{a}^{{mn}}}\\\therefore {{4}^{{2(2b+5)}}} & ={{4}^{{3(b+3)}}}\\\therefore {{4}^{{4b+10}}} & ={{4}^{{3b+9}}}\\\therefore 4b+10 & =3b+9\\\therefore b & =-1\end{align*}
You could also have changed the bases to $\scriptsize 2$.
3. .
\scriptsize \begin{align*}{{81}^{{x+3}}} & ={{27}^{{x-4}}}\\\therefore {{\left( {{{3}^{4}}} \right)}^{{x+3}}} & ={{\left( {{{3}^{3}}} \right)}^{{x-4}}}\\\therefore {{3}^{{4x+12}}} & ={{3}^{{3x-12}}}\\\therefore 4x+12 & =3x-12\\\therefore x & =-24\end{align*}
4. .
\scriptsize \begin{align*}{{81}^{y}}+{{9}^{{2y+1}}} & =270\\\therefore {{\left( {{{9}^{2}}} \right)}^{y}}+{{9}^{{2y+1}}} & =270\\\therefore {{9}^{{2y}}}+{{9}^{{2y}}}{{.9}^{1}} & =270\\\therefore {{9}^{{2y}}}+{{9.9}^{{2y}}} & =270\\\therefore {{10.9}^{{2y}}} & =270\\\therefore {{9}^{{2y}}} & =27\\\therefore {{\left( {{{3}^{2}}} \right)}^{{2y}}} & ={{3}^{3}}\\\therefore {{3}^{{4y}}} & ={{3}^{3}}\\\therefore 4y & =3\\\therefore y & =\displaystyle \frac{3}{4}\end{align*}
5. .
\scriptsize \begin{align*}{{2}^{x}}+{{2}^{{x+2}}} & =40\\\therefore {{2}^{x}}+{{2}^{x}}{{.2}^{2}} & =40\\\therefore {{2}^{x}}+{{4.2}^{x}} & =40\\\therefore {{5.2}^{x}} & =40\\\therefore {{2}^{x}} & =8\\\therefore {{2}^{x}} & ={{2}^{3}}\\\therefore x & =3\end{align*}

Back to Exercise 2.1

### Exercise 2.2

1. .
\scriptsize \begin{align*}2x+3y & =9\\\therefore 3y & =9-2x\\\therefore y & =\displaystyle \frac{{9-2x}}{3}\end{align*}
2. .
\scriptsize \begin{align*}s & =ut+\displaystyle \frac{1}{2}a{{t}^{2}}\\\therefore s-ut & =\displaystyle \frac{1}{2}a{{t}^{2}}\\\therefore 2(s-ut) & =a{{t}^{2}}\\\therefore \displaystyle \frac{{2s-2ut}}{{{{t}^{2}}}} & =a\\\therefore a & =\displaystyle \frac{{2s-2ut}}{{{{t}^{2}}}}\end{align*}
3. .
\scriptsize \begin{align*}A & =\pi {{R}^{2}}-\pi {{r}^{2}}\\\therefore \pi {{r}^{2}} & =\pi {{R}^{2}}-A\\\therefore {{r}^{2}} & =\displaystyle \frac{{\pi {{R}^{2}}-A}}{\pi }\\\therefore r & =\pm \sqrt{\frac{{\pi {{R}^{2}}-A}}{\pi }}}\end{align*
4. .
\scriptsize \begin{align*}r & =\sqrt{{{{a}^{2}}+{{b}^{2}}}}\\\therefore {{r}^{2}} & ={{\left( {\sqrt{{{{a}^{2}}+{{b}^{2}}}}} \right)}^{2}}\\\therefore {{r}^{2}} & ={{a}^{2}}+{{b}^{2}}\\\therefore {{b}^{2}} & ={{r}^{2}}-{{a}^{2}}\\\therefore b & =\pm \sqrt{{{{r}^{2}}-{{a}^{2}}}}\end{align*}
5. .
\scriptsize \begin{align*}{{F}_{R}} & =\displaystyle \frac{1}{2}\rho A{{C}_{D}}{{v}^{2}}\\\therefore 2{{F}_{R}} & =\rho A{{C}_{D}}{{v}^{2}}\\\therefore {{v}^{2}} & =\displaystyle \frac{{2{{F}_{R}}}}{{\rho A{{C}_{D}}}}\\\therefore v & =\pm \sqrt{\frac{{2{{F}_{R}}}}{{\rho A{{C}_{D}}}}}}\\\text{But }{{F}_{D}} & =24,\ A=0.5,\ {{C}_{D}}=0.01,\ \rho =1.2\\\therefore v & =\pm \sqrt{\frac{{2\times 24}}{{1.2\times 0.5\times 0.01}}}}\\\therefore v & =\pm \sqrt{{8\ 000}}\\\therefore v & =\pm 89.44\end{align*

Back to Exercise 2.2

### Unit 2: Assessment

1. Solve for the unknown in the following:
1. .
\scriptsize \begin{align*}{{12}^{{2x-3}}} & =1\\\therefore {{12}^{{2x-3}}} & ={{12}^{0}}\\\therefore 2x-3 & =0\\\therefore 2x & =3\\\therefore x & =\displaystyle \frac{3}{2}\end{align*}
2. .
\scriptsize \begin{align*}{{25}^{{1-2x}}}-{{5}^{4}} & =0\\\therefore {{\left( {{{5}^{2}}} \right)}^{{1-2x}}}-{{5}^{4}} & =0\\\therefore {{5}^{{2-4x}}}-{{5}^{4}} & =0\\\therefore {{5}^{{2-4x}}} & ={{5}^{4}}\\\therefore 2-4x & =4\\\therefore 4x & =-2\\\therefore x & =-\displaystyle \frac{1}{2}\end{align*}
3. .
\scriptsize \begin{align*}-\displaystyle \frac{1}{2}{{.6}^{\frac{m}{2}+3}}} & =-18\\\therefore {{6}^{\frac{m}{2}+3}}} & =36\\\therefore {{6}^{\frac{m}{2}+3}}} & ={{6}^{2}}\\\therefore \displaystyle \frac{m}{2}+3 & =2\\\therefore \displaystyle \frac{m}{2} & =-1\\\therefore m & =-2\end{align*
2. .
\scriptsize \begin{align*}D & =L+\displaystyle \frac{{{{W}^{2}}}}{{2L}}\\\therefore D-L & =\displaystyle \frac{{{{W}^{2}}}}{{2L}}\\\therefore {{W}^{2}} & =2L(D-L)\\\therefore W & =\pm \sqrt{{2L(D-L)}}\end{align*}
3. .
1. .
\scriptsize \begin{align*}V & =\displaystyle \frac{1}{3}\pi {{r}^{2}}h\\\therefore 3V & =\pi {{r}^{2}}h\\\therefore {{r}^{2}} & =\displaystyle \frac{{3V}}{{\pi h}}\\\therefore r & =\pm \sqrt{\frac{{3V}}{{\pi h}}}}\end{align*
2. .
\scriptsize \begin{align*}r & =\pm \sqrt{\frac{{3V}}{{\pi h}}}}\\&=\pm \sqrt{\frac{{3\times 250\ \text{m}{{\text{m}}^{\text{3}}}}}{{25\pi \ \text{mm}}}}}\\&=\pm 3.09\ \text{mm}\end{align*
However, as this is for the volume of a physical object, a negative answer does not make sense. Therefore $\scriptsize r=3.09\ \text{mm}$.

Back to Unit 1: Assessment 