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Exponentials and logarithmic functions

Composite functions

If we have two functions of $x$, $f$ and $g$, their sum, difference, product and quotient are also functions of $x$.

The sum and the difference of the two functions are

$(f+g)(x)=f(x)+g(x)$

and

$(f-g)(x)=f(x)-g(x)$

And, the product and the quotient of the two functions are

$(f\cdot g)(x)=f(x)\cdot g(x)$

and

$\left(\dfrac{f}{g}\right)(x)=\dfrac{f(x)}{g(x)}$, where $g(x)\ne 0$.

Example: Given, $f(x)=7x-2$ and $g(x)=11x+3$, find $(f+g)(x)$, $(f-g)(x)$, $(f\cdot g)(x)$ and $\left(\dfrac{f}{g}\right)(x)$.

Solution:

$(f+g)(x)=(7x-2)+(11x+3)$

Combining like terms,

$(f-g)(x)=18x+1$


$(f+g)(x)=(7x-2)-(11x+3)$

Combining like terms,

$(f+g)(x)=-4x-5$


$(f\cdot g)(x)=(7x-2)\cdot(11x+3)$

Multiplying and combining similar terms,

$(f\cdot g)(x)=77x^2-x-6$


$\left(\dfrac{f}{g}\right)(x)$$=\dfrac{7x-2}{11x+3}$

Composition of functions:

In general, we define a function as a function of a variable. That is, $f(x)$ is a function of the variable $x$, here $x$ is called the argument of the function. Sometimes, we may have the argument as another function, as $f(g(x))$, where $g(x)$ is another function. We call, $f(g(x))$ as the composition of $f$ and $g$. To represent a composition of functions, we use small circle between the functions.

i.e., The composition of $f$ and $g$ is denoted as $f\circ g$ and is defined as

$f\circ g=f(g(x))$

Example 1: Given $f(x)=x^2-7$ and $g(x)=\sqrt{x+10}$. Find $f\circ g$ and $g\circ f$.

Solution:

$(f\circ g)(x) =f(g(x))$

Take the function $f$ and replace the $x$ with $g$, i.e., $\sqrt{x+10}$:

$(f\circ g)(x) =(\sqrt{x+10})^2-7$

  $=x+10-7$

  $=x+3$


$(g\circ f)(x) =g(f(x))$

Take the function $g$ and replace the $x$ with $f$, i.e., $x^2-7$:

$(g\circ f)(x) =\sqrt{x^2-7+10}$

  $=\sqrt{x^2+3}$


Note that $f\circ g$ and $g\circ f$ are not equal.

Example 2: Given $f(x)=x^2+5$ and $g(x)=\sqrt{2x-8}$. Find $(f\circ g) (6)$ and $(g\circ f)(6)$.

Solution:

$(f\circ g)(6)=f(g(6))$

First find $g(6)$,

$g(6)=\sqrt{2\cdot 6 -8}$

  $=\sqrt{12-8}$

  $=\sqrt{4}$

  $=2$

Now, replace the $x$ in $f$ with this value,

$(f\circ g)(6)=f(2)$

  $=2^2+5$

  $=9$


$(g\circ f)(6)=g(f(6))$

First find $f(6)$,

$f(6)=6^2+5$

  $=36+5$

  $=41$

Now, replace the $x$ in $g$ with this value,

$(g\circ f)(6)=g(41)$

  $=\sqrt{2\cdot 42-8}$

  $=\sqrt{84-8}$

  $=\sqrt{76}$

  $=2\sqrt{19}$

Inverse functions

Logarithmic functions

If we find the inverse of an exponential function, we get another function, that we call a logarithmic function.

Consider an exponential equation,

$y=a^x$

And we want to solve this equation for $x$. To make it easy, we write the solution as a new function, called logarithmic function. It is written as

$x=\log_{a}y$

$a$ is called the base of the logarithm. That is same as the base of the exponent.

Here we have two equations, the first one is an exponential equation and the second one is the logarithmic form of the same equation. So, we can go from the exponential form to the logarithmic form and vice versa.

Example: Rewrite as a logarithmic equation, $3^4=81$

Solution:

First write the exponent, and put an equal sign and write the logarithm with the base and then the number on the other side:

$4=\log_{3}81$

or

$\log_{3}81=4$

Evaluating logarithmic expressions

We can evaluate a logarithmic expression by forming a logarithmic equation and writing it as an exponential equation.

Example 1: Evaluate the logarithmic expression, $\log_{8}64$.

Solution:

Take the expression as $x$,

$x=\log_{8}64$

Write this equation as an exponential equation,

$64=8^x$

We can write $64$ as $8^2$, therefore,

$8^2=8^x$

The bases are equal on both sides, so the power should be equal. Therefore,

$2=x$

or

$x=2$

Since $x$ is the given logarithmic expression, we have

$\boxed{\log_{8}64=2}$


Example 2: Evaluate the logarithmic expression, $\log_{4}\left(\dfrac{1}{64}\right)$.

Solution:

Properties of logarithm

$\log_{a}1=0$

$\log_{a}a=1$

$\log_{a}a^p=p$

$a^{\log_{a}x}=x$

Product property of logarithm

$\log_{a}{XY}=\log_{a}{X}+\log_{a}{Y}$

Quotient property of logarithm

$\log_{a}\left(\dfrac{X}{Y}\right)=\log_{a}{X}-\log_{a}{Y}$

Power property of logarithm

$\log_{a}x^p=p$


Example 1:Express the following as a single logarithm,

a. $\log_{7}5+\log_{7}11$

b. $\log_{6}13-\log_{6}4$

Solution:

a. $\log_{7}5+\log_{7}11=\log_{7}(5.11)$

  $=\log_{7}{55}$.

b. $\log_{6}13-\log_{6}4=\log_{6}\left(\dfrac{13}{4}\right)$.


Example 2: Expand the following logarithm using the properties of logarithms,

$\log \left(x^{10}y^3z\right)$.

Solution:

Use the product rule,

$\log \left(x^{10}y^3z\right)= \log x^{10}+\log y^3+\log z$.

  Apply the power rule,

  $=10\log x+3\log y+\log z$.

Example 3: Expand the following logarithm using the properties of logarithms,

$\log \left(\dfrac{x^{4}y^7}{z^6}\right)$.

Solution:

Use the product and the quotient rule,

$\log \left(\dfrac{x^{4}y^7}{z^6}\right)=\log x^{4}+\log y^7-\log z^6$.

  Apply the power rule,

  $=4\log x+7\log y-6\log z$.

Example 4: Write as a single logarithm,

$5\log_{b}x-4\left(\log_{b}u-3\log_{b}y\right)$

Solution:

Remove the parentheses,

$5\log_{b}x-4\left(\log_{b}u-3\log_{b}y\right)$

  $=5\log_{b}x-4\log_{b}u+12\log_{b}y$

Apply the power rule,

  $=\log_{b}x^5-\log_{b}u^4+\log_{b}y^{12}$

Apply the product and the quotient rule to combine,

  $=\log_{b}\left(\dfrac{x^5y^{12}}{u^4}\right)$

Compound interest problems

If you invest an amount, say $P$ at an annual interest rate $r$. Then the amount in your account after a certain time period is given by

$A(t)=P\left(1+\dfrac{r}{n}\right)^{nt}$

 where $t$ is the number of years, $n$ is the number of times per year the interest is compounded. $n=1$ if the interest is compounded annually and $n=12$ is if the interest is compounded monthly.

The interest can also be compounded continuously. If the interest is compounded continuously, then

$A(t)=Pe^{rt}$

Example 1:John deposited $\$3900$ in a bank account with an interest rate of $1.8\%$, compounded each year. How much money will be in his account after $6$ years? Assume that there is no withdrawals. Round your answer to the nearest cent.

Solution:

The original amount deposited is, $P=3900$.

Number of years, $n=6$

Interest rate is $r=1.8%=1.8/100=0.018$

Interest is compounded each year, i.e., annually, therefore, $n=1$.

The formula to find the amount at the end of the period is

$A(t)=P\left(1+\dfrac{r}{n}\right)^{nt}$

Substituting the values,

$A(t)=3900\left(1+\dfrac{0.018}{1}\right)^{1\cdot 6}$

  $=3900(1.018)^6$

  $=4340.62$

  $\boxed{Answer=\$ 4340.62}$

Changing the base of a logarithm

A logarithm with a base of $10$ is called the common logarithm. We can find the value of a common logarithm or a natural logarithm by using a calculator. But if the base is not $10$ or $e$, we need to change the base into one of these so that we can use the calculator.

If you have a logarithm with base $b$ and want to change the base into $a$, change the base into $a$ and divide that logarithm by the logarithm with base $a$ of $b$,

$\log_{b}x=\dfrac{\log_{a}x}{\log_{a}b}$


Example 1: Use the change of base formula to evaluate,

$\log_{4}78$

Round your answer to the nearest thousandth.

Solution:

Change the base from $4$ to $10$,

$\log_{4}78=\dfrac{\log_{10}78}{\log_{10}4}$

Use a calculator,

  $=\dfrac{1.8921}{0.6021}$

  $=3.143$

Example 2: Use the change of base formula to evaluate,

$\log_{2/7}\left(\dfrac{1}{11}\right)$

Round your answer to the nearest thousandth.

Solution:

Change the base to $10$,

$\log_{2/7}\left(\dfrac{1}{11}\right)=\dfrac{\log \left(\dfrac{1}{11}\right)}{\log \left(\dfrac{2}{7}\right)}$

Use a calculator,

  $=\dfrac{-1.0414}{-0.5441}$

  $=1.914$