02.5 Exponential Functions

6. Understanding exponential graphs

\[ y=3^x. \]

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7. Solving basic exponential equations

The main method in this lesson is to rewrite both sides using the same base.

For a valid exponential base:

\[ a^u=a^v \quad\Longrightarrow\quad u=v. \]

This works because the exponential function is strictly monotonic: different inputs give different outputs.

Worked example 4: A direct equation

Solve:

\[ 2^x=32. \]

Rewrite:

\[ 32=2^5. \]

Therefore:

\[ 2^x=2^5. \]

Equate the exponents:

\[ x=5. \]

Worked example 5: An expression in the exponent

Solve:

\[ 3^{2x-1}=27. \]

Rewrite:

\[ 27=3^3. \]

Then:

\[ 3^{2x-1}=3^3. \]

Therefore:

\[ 2x-1=3. \]\[ 2x=4. \]\[ x=2. \]

Check:

\[ 3^{2(2)-1}=3^3=27. \]

Worked example 6: A reciprocal

Solve:

\[ 5^{x+1}=\frac{1}{25}. \]

Rewrite:

\[ \frac{1}{25}=5^{-2}. \]

Therefore:

\[ x+1=-2. \]\[ x=-3. \]

Worked example 7: Rewriting both bases

Solve:

\[ 4^x=8. \]

Write both sides as powers of \(2\):

\[ (2^2)^x=2^3. \]

Apply the power-of-a-power rule:

\[ 2^{2x}=2^3. \]

Therefore:

\[ 2x=3. \]\[ x=\frac{3}{2}. \]

Worked example 8: Variable exponents on both sides

Solve:

\[ 9^x=3^{x+2}. \]

Rewrite:

\[ 9^x=(3^2)^x=3^{2x}. \]

Thus:

\[ 3^{2x}=3^{x+2}. \]

Equate exponents:

\[ 2x=x+2. \]

Therefore:

\[ x=2. \]

Equations with no real solution

Consider:

\[ 2^x=-4. \]

Since:

\[ 2^x>0 \]

for every real \(x\), this equation has no real solution.

Similarly:

\[ 2^x=0 \]

has no real solution.

Scope of this method: Some equations, such as \(2^x=7\), do have real solutions but cannot be solved by simple matching of familiar powers. Logarithms provide a general method and can be studied separately.

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8. Comparing exponential expressions

Choose a method based on whether the bases or exponents are the same.

8.1 Same base greater than \(1\)

If:

\[ a>1, \]

then a larger exponent gives a larger output.

For example:

\[ 3<5 \quad\Longrightarrow\quad 2^3<2^5. \]

This also applies to negative exponents:

\[ -4<-1 \quad\Longrightarrow\quad 3^{-4}<3^{-1}. \]

8.2 Same base between \(0\) and \(1\)

If:

\[ 0<a<1, \]

then a larger exponent gives a smaller output.

For example:

\[ 2<5 \]

implies:

\[ \left(\frac{1}{3}\right)^2 > \left(\frac{1}{3}\right)^5. \]

The order reverses because the function is decreasing.

Worked example 9

Compare:

\[ \left(\frac{1}{2}\right)^{-2} \quad\text{and}\quad \left(\frac{1}{2}\right)^3. \]

Since:

\[ -2<3, \]

and the base lies between zero and one:

\[ \left(\frac{1}{2}\right)^{-2} > \left(\frac{1}{2}\right)^3. \]

Checking:

\[ 4>\frac{1}{8}. \]

8.3 Different bases: Rewrite to a common base

Compare:

\[ 4^3 \quad\text{and}\quad 8^2. \]

Rewrite:

\[ 4^3=(2^2)^3=2^6, \]\[ 8^2=(2^3)^2=2^6. \]

Therefore:

\[ 4^3=8^2. \]

8.4 Same exponent, different positive bases

For a fixed positive exponent, a larger positive base gives a larger value:

\[ 2^3<5^3. \]

For a fixed negative exponent, the order reverses:

\[ 2^{-2}=\frac{1}{4}, \]\[ 5^{-2}=\frac{1}{25}. \]

Therefore:

\[ 2^{-2}>5^{-2}. \]

Do not compare expressions with both different bases and different exponents by looking at the bases alone.

For example:

\[ 2^5=32>27=3^3, \]

even though:

\[ 2<3. \]

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9. Simple exponential growth applications

Exponential growth occurs when a positive quantity increases by the same percentage over equal time intervals.

The model is:

\[ Q(t)=Q_0(1+r)^t. \]

Here:

• \(Q_0\) is the initial quantity.
• \(r\) is the growth rate per period, written as a decimal.
• \(t\) is the number of periods.
• \(1+r\) is the growth factor.

For a growth rate of \(10\%\):

\[ r=0.10, \]

so the growth factor is:

\[ 1+r=1.10. \]

Worked example 10: Annual growth

A quantity starts at \(1000\) units and grows by \(10\%\) each year. Find its value after three years, assuming the rate stays constant.

The model is:

\[ Q(t)=1000(1.10)^t. \]

After three years:

\[ Q(3)=1000(1.10)^3. \]\[ Q(3)=1000(1.331)=1331. \]

Therefore, the quantity is:

\[ 1331\text{ units}. \]

The increase is calculated from the updated amount each year:

\[ 1000\rightarrow 1100\rightarrow 1210\rightarrow 1331. \]

It is not a fixed increase of \(100\) units every year.

Worked example 11: Doubling

A laboratory culture initially contains \(200\) bacteria and doubles every hour under the model.

After \(t\) hours:

\[ N(t)=200\cdot 2^t. \]

After four hours:

\[ N(4)=200\cdot 2^4. \]\[ N(4)=200\cdot 16=3200. \]

The model predicts:

\[ 3200\text{ bacteria}. \]

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10. Simple exponential decay applications

Exponential decay occurs when a positive quantity decreases by the same percentage over equal time intervals.

The model is:

\[ Q(t)=Q_0(1-r)^t, \qquad 0<r<1. \]

The factor:

\[ 1-r \]

is the proportion remaining after each period.

For a decrease of \(20\%\):

\[ r=0.20. \]

The proportion remaining is:

\[ 1-0.20=0.80. \]

Worked example 12: Depreciation

A machine is initially worth \(5000\) RMB. Its value decreases by \(20\%\) each year.

Its value after \(t\) years is modeled by:

\[ V(t)=5000(0.80)^t. \]

After two years:

\[ V(2)=5000(0.80)^2. \]\[ V(2)=5000(0.64)=3200. \]

The predicted value is:

\[ 3200\text{ RMB}. \]

Important: A decrease of \(20\%\) means multiplying by \(0.80\), not \(0.20\).

Worked example 13: Repeated halving

A substance initially has a mass of \(80\) grams. Its mass halves every three hours.

The number of three-hour periods in \(t\) hours is:

\[ \frac{t}{3}. \]

Therefore:

\[ M(t)=80\left(\frac{1}{2}\right)^{t/3}. \]

After nine hours:

\[ M(9)=80\left(\frac{1}{2}\right)^{9/3}. \]\[ M(9)=80\left(\frac{1}{2}\right)^3. \]\[ M(9)=80\cdot\frac{1}{8}=10. \]

The remaining mass is:

\[ 10\text{ grams}. \]

Modeling reminder

Match the exponent to the stated time period. An annual rate requires time measured in years; a monthly rate requires time measured in months.

These models assume the stated percentage or multiplication factor stays constant over the modeled period.

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11. Common mistakes

Mistake 1: Multiplying the base by the exponent

Incorrect:

\[ 2^3=2\cdot 3. \]

Correct:

\[ 2^3=2\cdot 2\cdot 2=8. \]

Mistake 2: Making a negative exponent produce a negative output

Incorrect:

\[ 2^{-3}=-8. \]

Correct:

\[ 2^{-3}=\frac{1}{8}. \]

Mistake 3: Applying exponent laws to addition

In general:

\[ a^m+a^n\neq a^{m+n}. \]

For example:

\[ 2^2+2^3=4+8=12, \]

whereas:

\[ 2^{2+3}=32. \]

The exponent-addition rule applies to multiplication of powers with the same base.

Mistake 4: Equating exponents before matching bases

From:

\[ 4^x=8, \]

you cannot conclude:

\[ x=1. \]

First rewrite:

\[ 2^{2x}=2^3. \]

Then:

\[ x=\frac{3}{2}. \]

Mistake 5: Forgetting that fractional bases reverse comparisons

Since:

\[ 0<\frac{1}{2}<1, \]

we have:

\[ \left(\frac{1}{2}\right)^5 < \left(\frac{1}{2}\right)^2. \]

Mistake 6: Using the percentage lost as the remaining factor

A loss of \(15\%\) gives the remaining factor:

\[ 1-0.15=0.85. \]

Use:

\[ Q(t)=Q_0(0.85)^t. \]

Mistake 7: Assuming repeated growth and decay cancel

An increase of \(10\%\), followed by a decrease of \(10\%\), multiplies the original quantity by:

\[ 1.10\cdot 0.90=0.99. \]

The final quantity is \(99\%\) of the original, not \(100\%\).

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12. Practice quiz

Try these questions before reading the answers.

Question 1: Recognizing exponential functions

Which are exponential functions in the basic form studied in this lesson?

A.

\[ f(x)=5^x \]

B.

\[ g(x)=x^5 \]

C.

\[ h(x)=\left(\frac{1}{4}\right)^x \]

D.

\[ p(x)=(-3)^x \]

Question 2: Evaluation

Evaluate:

\[ 4^{-2}. \]

Question 3: Fractional exponents

Evaluate:

\[ 27^{2/3}. \]

Question 4: Simplifying

Simplify:

\[ \frac{3^{x+4}}{3^{x+1}}. \]

Question 5: Function properties

For:

\[ f(x)=\left(\frac{1}{5}\right)^x, \]

state:

• The domain.
• The range.
• Whether the function increases or decreases.
• The \(y\)-intercept.

Question 6: Identifying a graph

A basic exponential graph passes through:

\[ (0,1) \quad\text{and}\quad (1,4). \]

Find its equation.

Question 7: Solving an equation

Solve:

\[ 2^{x+2}=32. \]

Question 8: Matching bases

Solve:

\[ 9^x=27. \]

Question 9: Comparing expressions

Compare:

\[ \left(\frac{1}{3}\right)^2 \quad\text{and}\quad \left(\frac{1}{3}\right)^5. \]

Question 10: Comparing different bases

Compare:

\[ 2^6 \quad\text{and}\quad 4^3. \]

Question 11: Growth

A quantity starts at \(500\) units and grows by \(20\%\) per year.

Find its value after two years.

Question 12: Decay

An item initially costs \(8000\) RMB and loses \(25\%\) of its value each year.

Find its value after two years.

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13. Answers and explanations

Answer 1

A and C.

Both have a fixed positive base different from \(1\), with the variable in the exponent.

Option B is a power function.

Option D does not define a real-valued exponential function for every real input because its base is negative.

Answer 2

\[ 4^{-2}=\frac{1}{4^2}=\frac{1}{16}. \]

Answer 3

\[ 27^{2/3} =\left(\sqrt[3]{27}\right)^2 =3^2 =9. \]

Answer 4

\[ \frac{3^{x+4}}{3^{x+1}} =3^{(x+4)-(x+1)}. \]\[ =3^3=27. \]

Answer 5

Domain:

\[ (-\infty,\infty). \]

Range:

\[ (0,\infty). \]

The function is strictly decreasing because:

\[ 0<\frac{1}{5}<1. \]

The \(y\)-intercept is:

\[ (0,1). \]

Answer 6

For:

\[ f(x)=a^x, \]

we have:

\[ f(1)=a. \]

Since the graph passes through \((1,4)\):

\[ a=4. \]

Therefore:

\[ f(x)=4^x. \]

Answer 7

\[ 2^{x+2}=32=2^5. \]

Therefore:

\[ x+2=5, \]\[ x=3. \]

Answer 8

Rewrite both sides using base \(3\):

\[ (3^2)^x=3^3. \]\[ 3^{2x}=3^3. \]

Therefore:

\[ 2x=3, \]\[ x=\frac{3}{2}. \]

Answer 9

The base is between zero and one, so the function is decreasing.

Since:

\[ 2<5, \]

we obtain:

\[ \left(\frac{1}{3}\right)^2 > \left(\frac{1}{3}\right)^5. \]

Answer 10

\[ 4^3=(2^2)^3=2^6. \]

Therefore:

\[ 2^6=4^3. \]

Answer 11

The growth factor is:

\[ 1+0.20=1.20. \]

After two years:

\[ Q(2)=500(1.20)^2. \]\[ Q(2)=500(1.44)=720. \]

The quantity is:

\[ 720\text{ units}. \]

Answer 12

The remaining factor is:

\[ 1-0.25=0.75. \]

After two years:

\[ V(2)=8000(0.75)^2. \]\[ V(2)=8000(0.5625)=4500. \]

The predicted value is:

\[ 4500\text{ RMB}. \]

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14. Lesson review

You should now be able to:

• Recognize the basic form of an exponential function:

\[ f(x)=a^x, \qquad a>0, \qquad a\neq 1. \]

• Evaluate expressions using zero, negative, and fractional exponents.
• Identify growth when \(a>1\) and decay when \(0<a<1\).
• State the domain and range:

\[ \text{Domain}=(-\infty,\infty), \qquad \text{Range}=(0,\infty). \]

• Recognize the point \((0,1)\) and the horizontal asymptote \(y=0\).
• Solve elementary equations by rewriting both sides with the same base.
• Compare exponential expressions using monotonicity.
• Apply constant-percentage growth and decay models:

\[ Q(t)=Q_0(1+r)^t, \]\[ Q(t)=Q_0(1-r)^t. \]