Techniques for Integration: Background You’ll Need 2

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Techniques for Integration: Background You’ll Need 2

Apply properties of exponential and logarithmic functions

Exponential Functions

Exponential functions arise in many applications. One common example is population growth.

If a population starts with $P_0$ individuals and then grows at an annual rate of $2\%$ , its population after $1$ year is

P (1) = P_{0} + 0.02 P_{0} = P_{0} (1 + 0.02) = P_{0} (1.02)

$P(1)=P_0+0.02P_0=P_0(1+0.02)=P_0(1.02)$

Its population after $2$ years is

P (2) = P (1) + 0.02 P (1) = P (1) (1.02) = P_{0} (1.02)^{2}

$P(2)=P(1)+0.02P(1)=P(1)(1.02)=P_0(1.02)^2$

In general, its population after $t$ years is

P (t) = P_{0} (1.02)^{t}

$P(t)=P_0(1.02)^t$ ,

which is an exponential function.

More generally, any function of the form $f(x)=b^x$ , where $b>0, \, b \ne 1$ , is an exponential function with base $b$ and exponent $x$ . Exponential functions have constant bases and variable exponents.

exponential function

For any real number $x$ , an exponential function is a function with the form

$f(x)=ab^x$

where,

$a$ is a non-zero real number called the initial value and
$b$ is any positive real number ( $b>0$ ) such that $b≠1$ .

Evaluating Exponential Functions

To evaluate an exponential function with the form $f(x)=b^x$ , we simply substitute $x$ with the given value, and calculate the resulting power.

Let $f(x)=2^x$ . What is $f(3)$ ?

\begin{array}{rcl} f (x) & = & 2^{x} \\ f (3) & = & 2^{3} & Substitute x = 3. \\ = & 8 & Evaluate the power. \end{array}

$\begin{array}{rcl} f(x) & = & 2^x \\ f(3) & = & 2^3 & \quad \text{Substitute } x = 3. \\ & = & 8 & \quad \text{Evaluate the power.} \end{array}$

To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations.

Let $f(x)=30(2)^x$ . What is $f(3)$ ?

\begin{array}{rcll} f (x) & = & 30 (2)^{x} \\ f (3) & = & 30 (2)^{3} & Substitute x = 3. \\ = & 30 (8) & Simplify the power first. \\ = & 240 & Multiply. \end{array}

$\begin{array}{rcll} f(x) & = & 30(2)^x & \\ f(3) & = & 30(2)^3 & \quad \text{Substitute } x = 3. \\ & = & 30(8) & \quad \text{Simplify the power first.} \\ & = & 240 & \quad \text{Multiply.} \end{array}$

Note that if the order of operations were not followed, the result would be incorrect:

f (3) = 30 (2)^{3} \neq 60^{3} = 216, 000

$f(3)=30(2)^3≠60^3=216,000$

How To: Evaluating Exponential Functions

Given an exponential function, identify $a$ , $b$ , and the value of $x$ you’re being asked to substitute into the function.
Replace the variable $x$ in the function with the given number.
Compute the value of $b^x$ . This means raising the base $b$ to the power of $x$ .
If there is a coefficient $a$ in front of the base, multiply the result of $b^x$ by $a$ . If $a$ is $1$ , this step does not change the value.
Simplify the expression if necessary. This could involve performing any additional multiplication or addition/subtraction if the function has more terms.

Let $f(x)=5(3)^x+1$ . Evaluate $f(2)$ without using a calculator.

Follow the order of operations. Be sure to pay attention to the parentheses.

\begin{array}{rcll} f (x) & = & 5 (3)^{x + 1} \\ f (2) & = & 5 (3)^{2 + 1} & Substitute x = 2. \\ = & 5 (3)^{3} & Add the exponents. \\ = & 5 (27) & Simplify the power. \\ = & 135 & Multiply. \end{array}

$\begin{array}{rcll} f(x) & = & 5(3)^{x+1} & \\ f(2) & = & 5(3)^{2+1} & \quad \text{Substitute } x = 2. \\ & = & 5(3)^3 & \quad \text{Add the exponents.} \\ & = & 5(27) & \quad \text{Simplify the power.} \\ & = & 135 & \quad \text{Multiply.} \end{array}$

Suppose a particular population of bacteria is known to double in size every $4$ hours. If a culture starts with $1000$ bacteria, the number of bacteria after $4$ hours is $n(4)=1000·2$ . The number of bacteria after $8$ hours is $n(8)=n(4)·2=1000·2^2$ .

In general, the number of bacteria after $4m$ hours is $n(4m)=1000·2^m$ . Letting $t=4m$ , we see that the number of bacteria after $t$ hours is $n(t)=1000·2^{t/4}$ .

Find the number of bacteria after $6$ hours, $10$ hours, and $24$ hours.

The number of bacteria after $6$ hours is given by $n(6)=1000·2^{6/4} \approx 2828$ bacteria.

The number of bacteria after $10$ hours is given by $n(10)=1000·2^{10/4} \approx 5657$ bacteria.

The number of bacteria after $24$ hours is given by $n(24)=1000·2^{24/4}=1000·2^6=64,000$ bacteria.

Laws of Exponents

The Laws of Exponents are fundamental rules that govern the operations involving powers. These rules are essential for simplifying expressions and are foundational for higher-level math.

laws of exponents

The Product of Powers rule states that when you multiply two exponents with the same base, you can add the exponents.
$b^x·b^y=b^{x+y}$
The Quotient of Powers rule tells us that when dividing exponents with the same base, we subtract the exponents.
$\large\frac{b^x}{b^y} \normalsize = b^{x-y}$
The Power of a Power rule shows that when taking an exponent to another exponent, we multiply the exponents.
$(b^x)^y=b^{xy}$
The Power of a Product rule lets us know that when raising a product to an exponent, each factor in the product is raised to the exponent.
$(ab)^x=a^x b^x$
The Power of a Quotient rule indicates that when a quotient is raised to an exponent, both the numerator and the denominator are raised to the exponent.
$\dfrac{a^x}{b^x} =\left(\dfrac{a}{b}\right)^x$

Note: This is true for any constants $a>0, \, b>0$ , and for all $x$ and $y$

Use the laws of exponents to simplify each of the following expressions.

$\large \frac{(2x^{2/3})^3}{(4x^{-1/3})^2}$
$\large \frac{(x^3 y^{-1})^2}{(xy^2)^{-2}}$

We can simplify as follows:
$\large \frac{(2x^{2/3})^3}{(4x^{-1/3})^2} \normalsize = \large \frac{2^3(x^{2/3})^3}{4^2(x^{-1/3})^2} \normalsize = \large \frac{8x^2}{16x^{-2/3}} \normalsize = \large \frac{x^2x^{2/3}}{2} \normalsize = \large \frac{x^{8/3}}{2}$
We can simplify as follows:
$\large \frac{(x^3y^{-1})^2}{(xy^2)^{-2}} \normalsize = \large \frac{(x^3)^2(y^{-1})^2}{x^{-2}(y^2)^{-2}} \normalsize = \large \frac{x^6y^{-2}}{x^{-2}y^{-4}} \normalsize = x^6x^2y^{-2}y^4 = x^8y^2$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this video using this link (opens in new window).

When you encounter a negative exponent on a term in the denominator of a fraction, you can transform it into a positive exponent by moving the term to the numerator.

\frac{1}{a^{-} n} = a^{n}

$\frac{1}{a^-n}=a^{n}$

Using this rule can significantly simplify expressions involving exponents.

Logarithmic Functions

Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions.

Inverse Functions

For any one-to-one function $f\left(x\right)=y$ , a function ${f}^{-1}\left(x\right)$ is an inverse function of $f$ if ${f}^{-1}\left(y\right)=x$ .

The notation ${f}^{-1}$ is read “ $f$ inverse.” Like any other function, we can use any variable name as the input for ${f}^{-1}$ , so we will often write ${f}^{-1}\left(x\right)$ , which we read as $"f$ inverse of $x$ “.

Logarithmic functions come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.

The exponential function $f(x)=b^x$ is one-to-one, with domain $(−\infty ,\infty)$ and range $(0,\infty )$ . Therefore, it has an inverse function, called the logarithmic function with base $b$ .

For any $b>0, \, b \ne 1$ , the logarithmic function with base $b$ , denoted $\log_b$ , has domain $(0,\infty )$ and range $(−\infty ,\infty )$ , and satisfies

$\log_b(x)=y$ if and only if $b^y=x$ .

logarithmic functions

A logarithmic function is the inverse of an exponential function and is written as $log_{b}(x)$ . For a given base $b$ , it tells us the power to which $b$ must be raised to get $x$ .

\begin{array}{cccc} \log_{2} (8) = 3 & since 2^{3} = 8, \\ \log_{10} (\frac{1}{100}) = - 2 & since 10^{- 2} = \frac{1}{10^{2}} = \frac{1}{100}, \\ \log_{b} (1) = 0 & since b^{0} = 1 for any base b > 0. \end{array}

$\begin{array}{cccc} \log_2 (8)=3\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}2^3=8,\hfill \\ \log_{10} (\frac{1}{100})=-2\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}10^{-2}=\frac{1}{10^2}=\frac{1}{100},\hfill \\ \log_b (1)=0\hfill & & & \text{since}\phantom{\rule{3em}{0ex}}b^0=1 \, \text{for any base} \, b>0.\hfill \end{array}$

The most commonly used logarithmic function is the function $\log_e (x)$ . Since this function uses natural $e$ as its base, it is called the natural logarithm. Here we use the notation $\ln(x)$ or $\ln x$ to mean $\log_e (x)$ .

\begin{array}{l} \ln (e) = \log_{e} (e) = 1 \\ \ln (e^{3}) = \log_{e} (e^{3}) = 3 \\ \ln (1) = \log_{e} (1) = 0 \end{array}

$\begin{array}{l}\ln (e)=\log_e (e)=1 \\ \ln(e^3)=\log_e (e^3)=3 \\ \ln(1)=\log_e (1)=0\end{array}$

Euler’s number, denoted as $e$ , is a fundamental mathematical constant approximately equal to $2.71828$ . It is the base of the natural logarithm and the natural exponential function, known for its unique properties in calculus, especially in relation to growth processes and compound interest calculations.

Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.

Properties of Logarithms

If $a,b,c>0, \, b\ne 1$ , and $r$ is any real number, then

$\begin{array}{cccc}1.\phantom{\rule{2em}{0ex}}\log_b (ac)=\log_b (a)+\log_b (c)\hfill & & & \text{(Product property)}\hfill \\ 2.\phantom{\rule{2em}{0ex}}\log_b(\frac{a}{c})=\log_b (a) -\log_b (c)\hfill & & & \text{(Quotient property)}\hfill \\ 3.\phantom{\rule{2em}{0ex}}\log_b (a^r)=r \log_b (a)\hfill & & & \text{(Power property)}\hfill \end{array}$

Solve each of the following equations for $x$ .

$\ln \left(\frac{1}{x}\right)=4$
$\log_{10} \sqrt{x}+ \log_{10} x=2$
$\ln(2x)-3 \ln(x^2)=0$

By the definition of the natural logarithm function,
$\ln\big(\frac{1}{x}\big)=4 \, \text{ if and only if } \, e^4=\frac{1}{x}$

Therefore, the solution is $x=\frac{1}{e^4}$ .
Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation as
$\log_{10} \sqrt{x}+ \log_{10} x = \log_{10} x \sqrt{x} = \log_{10}x^{3/2} = \frac{3}{2} \log_{10} x$

Therefore, the equation can be rewritten as

$\frac{3}{2} \log_{10} x = 2 \, \text{ or } \, \log_{10} x = \frac{4}{3}$

The solution is $x=10^{4/3}=10\sqrt[3]{10}$ .
Using the power property of logarithmic functions, we can rewrite the equation as $\ln(2x) - \ln(x^6) = 0$ .
Using the quotient property, this becomes

$\ln\big(\frac{2}{x^5}\big)=0$

Therefore, $\frac{2}{x^5}=1$ , which implies $x=\sqrt[5]{2}$ . We should then check for any extraneous solutions.

Solve each of the following equations for $x$ .

$5^x=2$
$e^x+6e^{−x}=5$

Applying the natural logarithm function to both sides of the equation, we have
$\ln 5^x=\ln 2$

Using the power property of logarithms,

$x \ln 5=\ln 2$

Therefore, $x=\frac{\ln 2 }{\ln 5}$ .
Multiplying both sides of the equation by $e^x$ , we arrive at the equation
$e^{2x}+6=5e^x$

Rewriting this equation as

$e^{2x}-5e^x+6=0$ ,

we can then rewrite it as a quadratic equation in $e^x$ :

$(e^x)^2-5(e^x)+6=0$

Now we can solve the quadratic equation. Factoring this equation, we obtain

$(e^x-3)(e^x-2)=0$

Therefore, the solutions satisfy $e^x=3$ and $e^x=2$ . Taking the natural logarithm of both sides gives us the solutions

$x=\ln 3, \, \ln 2$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this video using this link (opens in new window).