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Section 10.4 Exponential Functions

Objective: Solve exponential equations by finding a common base.

As our study of algebra gets more advanced, we begin to study more involved functions. One pair of inverse functions we will look at are exponential functions and logarithmic functions. Here we will look at exponential functions and then we will consider logarithmic functions in another lesson. Exponential functions are functions where the variable is in the exponent such as \(f(x) = a^x\text{.}\) (It is important not to confuse exponential functions with polynomial functions where the variable is in the base such as \(f(x)=x^2\)).

World View Note: One common application of exponential functions is population growth. According to the \(2009\) CIA World Factbook, the country with the highest population growth rate is a tie between the United Arab Emirates (north of Saudi Arabia) and Burundi (central Africa) at \(3.69\%\text{.}\) There are \(32\) countries with negative growth rates, the lowest being the Northern Mariana Islands (north of Australia) at \(-7.08\%\text{.}\)

Solving exponetial equations cannot be done using the skill set we have seen in the past. For example, if \(3x = 9\text{,}\) we cannot take the \(x\) - root of \(9\) because we do not know what the index is and this doesn’t get us any closer to finding \(x\text{.}\) However, we may notice that \(9\) is \(32\text{.}\) We can then conclude that if \(3x = 32\) then \(x = 2\text{.}\) This is the process we will use to solve exponential functions. If we can re-write a problem so the bases match, then the exponents must also match.

Example 10.4.1.

\begin{align*} 5^{2x+1}=125 \amp \quad \text{Rewrite $125$ as } 5^3 \\ 5^{2x+1}=5^3 \amp\quad \text{Same base, set exponents equal} \\ 2x+1=3 \amp\quad \text{Solve} \\ \underline{-1-1} \amp\quad \text{Subtract $1$ from both sides} \\ 2x=2 \amp\quad \text{Divide both sides by } 2 \\ \overline{2} \qquad \overline{2} \amp\quad \text{ } \\ x=1 \amp\quad \text{ Our Solution } \end{align*}

Sometimes we may have to do work on both sides of the equation to get a common base. As we do so, we will use various exponent properties to help. First we will use the exponent property that states \((a^x)^y =a^{xy}\text{.}\)

Example 10.4.2.

\begin{align*} 8^{3x}=32 \amp \quad \text{Rewrite $8$ as $2^3$ and $32$ as } 2^5 \\ (2^3)^{3x}=2^5 \amp\quad \text{Multiply exponents $3$ and } 3x \\ 2^{9x}=2^5 \amp\quad \text{Same base, set exponents equal} \\ 9x=5 \amp\quad \text{Solve} \\ \overline{9} \qquad \overline{9} \amp\quad \text{Divide both sides by } 9 \\ x=\frac{5}{9} \amp\quad \text{ Our Solution } \end{align*}

As we multiply exponents we may need to distribute if there several terms involved.

Example 10.4.3.

\begin{align*} 27^{3x+5}=81^{4x+1} \amp \quad \text{Rewrite $125$ as } 5^3 \\ \left(3^3\right)^{3x+5}=(3^4)^{4x+1} \amp\quad \text{} \\ 3^{9x+15}=3^{16x+4} \amp\quad \text{Same base, set exponents equal} \\ 9x+15=16x+4 \amp\quad \text{Move variables to one side} \\ \underline{-9x \quad\quad -9x \qquad} \amp\quad \text{Subtract $9x$ from both sides } \\ 15=7x+4 \amp\quad \text{Subtract $4$ from both sides } \\ \underline{-4 \qquad \ -4} \amp\quad \text{ } \\ 11=7x \amp\quad \text{ Divide both sides by } 7 \\ \overline{7} \qquad \overline{7} \amp\quad \text{ } \\ \frac{11}{7}=x \amp\quad \text{ Our Solution } \end{align*}

Another useful exponent property is that negative exponents will give us a reciprocal, \(\frac{1}{a^n}=a^{-n}\)

Example 10.4.4.

\begin{align*} \left( \frac{1}{9}\right)^{2x}=3^{7x-1} \amp \quad \text{Rewrite $\frac{1}{9}$ as $3^{-2}$ (negative exponent to flip) } \\ \left(3^{-2}\right)^{2x}=3^{7x-1} \amp\quad \text{Multiply exponents $-2$ and } 2x \\ 3^{-4x}=3^{7x-1} \amp\quad \text{Same base, set exponents equal} \\ -4x=7x-1 \amp\quad \text{Subtract $7x$ from both sides} \\ \underline{-7x -7x } \qquad \amp\quad \text{ } \\ -11x=-1 \amp\quad \text{Divide by } -11 \\ \overline{-11} \quad \overline{-11} \amp\quad \text{ } \\ x=\frac{1}{11} \amp\quad \text{ Our Solution } \end{align*}

If we have several factors with the same base on one side of the equation we can add the exponents using the property that states \(a^xa^y =a^{x+y}\text{.}\)

Example 10.4.5.

\begin{align*} 5^{4x} \cdot 5^{2x-1}=5^{3x+11} \amp \quad \text{Add exponents on left, combining like terms } \\ 5^{6x-1}=5^{3x+11} \amp\quad \text{Same base, set exponents equal } \\ 6x-1=3x+11 \amp\quad \text{Move variables to one sides} \\ \underline{-3x \qquad -3x} \qquad \amp\quad \text{Subtract $3x$ from both sides} \\ 3x-1=11 \amp\quad \text{Add $1$ to both sides } \\ \underline{+1 \ +1} \amp\quad \text{ } \\ 3x=12 \amp\quad \text{ Divide both sides by } 3 \\ \overline{3} \qquad \overline{3} \amp\quad \text{ } \\ x=4 \amp\quad \text{ Our Solution } \end{align*}

It may take a bit of practice to get use to knowing which base to use, but as we practice, we will get much quicker at knowing which base to use. As we do so, we will use our exponent properties to help us simplify. Again, below are the properties we used to simplify.

\begin{equation*} \mathbf{ \left(a^x\right)^y=a^{xy}} \ \ \textbf{ and } \ \ \mathbf{ \frac{1}{a^n}=a^{-n}} \ \ \textbf{ and } \ \ \mathbf{a^xa^y=a^{x+y}} \end{equation*}

We could see all three properties used in the same problem as we get a common base. This is shown in the next example.

Example 10.4.6.

\begin{align*} 16^{2x-5} \cdot \left(\frac{1}{4}\right)^{3x+1}=32 \cdot \left(\frac{1}{2}\right)^{x+3} \amp \quad \text{Write with a common base of } 2 \\ \left(2^4\right)^{2x-5} \cdot \left(2^{-2}\right)^{3x+1}=2^5 \cdot \left(2^{-1}\right)^{x+3} \amp\quad \text{ Multiply exponents, distributing as needed } \\ 2^{8x-20} \cdot 2^{-6x-2}=2^5 \cdot 2^{-x-3} \amp\quad \text{Add exponents, combining like terms } \\ 2^{2x-22}=2^{-x+2} \amp\quad \text{Same base, set exponents equal} \\ 2x-22=-x+2 \amp\quad \text{Move variables to one side } \\ \underline{+x \quad\quad\quad +x} \qquad \amp\quad \text{ Add $x$ to both sides } \\ 3x-22=2 \amp\quad \text{ Add $22$ to both sides } \\ \underline{+22+22} \amp\quad \text{ Divide both sides by } 3 \\ 3x=24 \amp\quad \text{ Divide both sides by } 3 \\ \overline{\ 3\ } \quad \overline{\ 3\ } \amp\quad \text{ } \\ x=8 \amp\quad \text{ Our Solution } \end{align*}

All the problems we have solved here we were able to write with a common base. However, not all problems can be written with a common base, for example, \(2 = 10^x\text{,}\) we cannot write this problem with a common base. To solve problems like this we will need to use the inverse of an exponential function. The inverse is called a logarithmic function, which we will discuss in another section.

Exercises Exercises – Exponential Functions

Exercise Group.

Solve each equation.

1.

\(3^{1-2n}=3^{1-3n} \)

2.

\(4^{2x}=\frac{1}{16} \)

3.

\(4^{2a}=1 \)

4.

\(16^{-3p}=64^{-3p} \)

5.

\(\left(\frac{1}{25}\right)^{-k}=125^{-2k-2} \)

6.

\(625^{-n-2}=\frac{1}{25} \)

7.

\(6^{2m+1}=\frac{1}{36} \)

8.

\(6^{2r-3}=6^{r-3} \)

9.

\(6^{-3x}=36 \)

10.

\(5^{2n}=5^{-n} \)

11.

\(64^b=2^5 \)

12.

\(216^{-3v}=36^{3v} \)

13.

\(\left(\frac{1}{4}\right)^x=16 \)

14.

\(27^{-2n-1}=9 \)

15.

\(4^{3a}=4^3 \)

16.

\(4^{-3v}=64 \)

17.

\(36^{3x}=216^{2x+1} \)

18.

\(64^{x+2}=16 \)

19.

\(9^{2n+3}=243 \)

20.

\(16^{2k}=\frac{1}{64} \)

21.

\(3^{3x-2}=3^{3x+1} \)

22.

\(243^p=27^{-3p} \)

23.

\(3^{-2x}=3^3 \)

24.

\(4^{2n}=4^{2-3n} \)

25.

\(5^{m+2} = 5^{-n} \)

26.

\(625^{2x}=25 \)

27.

\(\left(\frac{1}{36}\right)^{b-1}=216 \)

28.

\(216^{2n}=36 \)

29.

\(6^{2-2x}=6^2 \)

30.

\(\left(\frac{1}{4}\right)^{3v-2}=64^{1-v} \)

31.

\(4 \cdot 2^{-3n-1}=\frac{1}{4} \)

32.

\(\frac{216}{6^{-2a}}=6^{3a} \)

33.

\(4^{3k-3} \cdot 4^{2-2k} = 16^{-k} \)

34.

\(32^{2p-2} \cdot 8^p=\left(\frac{1}{2}\right)^{2p} \)

35.

\(9^{-2x} \cdot \left(\frac{1}{243}\right)^{3x}=243^{-x} \)

36.

\(3^{2m} \cdot 3^{3m}=1 \)

37.

\(64^{n-2} \cdot 16^{n+2} = \left(\frac{1}{4}\right)^{3n-1} \)

38.

\(3^{2-x} \cdot 3^{3m}=1 \)

39.

\(5^{-3n-3} \cdot 5^{2n} =1 \)

40.

\(4^{3r} \cdot 4^{-3r} = \frac{1}{64} \)