6. Exponential & Logarithmic Functions

Lesson

We have learned about exponents (indices) and ways we can manipulate them in math. In math, we have pairs of inverse functions, or opposite operations, i.e., addition & subtraction, multiplication & division. The inverse function of an **exponential** is called a logarithm, and it works like this:

Logarithmic form

If we have an exponential of the form:

$b^x=a$`b``x`=`a`

Then we can rewrite it as the logarithm:

$\log_ba=x$`l``o``g``b``a`=`x`

and $b$`b` is called the base, just like in exponentials

In other words, we use logarithms when we are interested in finding out the exponent needed ($x$`x`) to raise a certain base ($b$`b`) to a certain number ($a$`a`).

Careful!

An important part to remember is that you can only take the logarithm of a positive number with a positive base that's not one:

$\log_ba$`l``o``g``b``a` only makes sense when $a>0$`a`>0 and $b>0$`b`>0, $b\ne1$`b`≠1

Express $6^2=36$62=36 in logarithmic form

**Think: **Remember, just as in all inverse operations, we want the exponent on its own and the logarithm on the other side.

**Do:**

$6$6 is the base and $2$2 is the exponent, so:

$\log_636=2$`l``o``g`636=2

Rewrite in exponential form: $\log_432=2.5$`l``o``g`432=2.5

**Think: **Remember that in exponential form we want isolate the resulting number after having the base raised to the exponent.

**Do:** Identify the components: $base=4$`b``a``s``e`=4 and the result of the logarithm is $exponent=2.5$`e``x``p``o``n``e``n``t`=2.5 . So,

$4^{2.5}=32$42.5=32

Rewrite the equation $9^{\frac{3}{2}}=27$932=27 in logarithmic form.

**Think:**** **Remember, we want the exponent on its own and the logarithm on the other side.

**Do:** Identify the components: $base=9$`b``a``s``e`=9 and the exponent turns into the result $exponent=\frac{3}{2}$`e``x``p``o``n``e``n``t`=32 . So,

$\log_927=\frac{3}{2}$`l``o``g`927=32

Rewrite the equation $8^{-3}=\frac{1}{512}$8−3=1512 in logarithmic form

**Think:** Remember, we want the exponent on its own and the logarithm on the other side.

**Do:** Identify the components: $base=8$`b``a``s``e`=8 and the exponent turns into the result $exponent=-3$`e``x``p``o``n``e``n``t`=−3 . So,

$\log_8\frac{1}{512}=-3$`l``o``g`81512=−3

Evaluate $\log_216$`l``o``g`216.

Rewrite the equation $\left(0.9\right)^0=1$(0.9)0=1 in logarithmic form.

Rewrite the equation $9^{\frac{3}{2}}=27$932=27 in logarithmic form.

Recall that a relation and its inverse form mirror images of each other across the line $y=x$`y`=`x`.

The inverse of the logarithms function $y=\log_2\left(x-1\right)$`y`=`l``o``g`2(`x`−1) is the exponential function $y=2^x+1$`y`=2`x`+1. Note that the domain and range of the inverse function is the range and domain of the function respectively. The asymptotes are likewise reflections across the line $y=x$`y`=`x`.

When viewing the $x$`x` and $y$`y` tables for the logarithmic and exponential functions,

Exponential

$x$x |
$y=2^x+1$y=2x+1 |
---|---|

$0$0 | $2$2 |

$1$1 | $3$3 |

$2$2 | $5$5 |

$3$3 | $9$9 |

Logarithmic

$x$x |
$y=\log_2\left(x-1\right)$y=log2(x−1) |
---|---|

$2$2 | $0$0 |

$3$3 | $1$1 |

$5$5 | $2$2 |

$9$9 | $3$3 |

we notice that the ordered pairs switch positions - the $y$`y` elements move into the first position and the $x$`x` elements move into the second position of each ordered pair.

What is the inverse of $y=3^{x-2}-5$`y`=3`x`−2−5?

**Think:** Before finding the inverse, we must isolate what we are to un-do - exponent. After we isolate the exponent, we need to identify the components of the equation.

**Do:** We rewrite $y+5=3^{x-2}$`y`+5=3`x`−2, and again from the definition, $x-2=\log_3\left(y+5\right)$`x`−2=`l``o``g`3(`y`+5).

Thus the inverse becomes:

$y-2=\log_3\left(x+5\right)$`y`−2=`l``o``g`3(`x`+5)

$y=\log_3\left(x+5\right)+2$`y`=`l``o``g`3(`x`+5)+2

Find the inverse of $f\left(x\right)=2\log_e\left(x+1\right)-3$`f`(`x`)=2`l``o``g``e`(`x`+1)−3 .

**Think: **Finding the inverse of a natural logarithmic function can be similarly found as the previous examples. Before we start, it is best to isolate the operation we want to un-do: natural logarithm.

**Do: **Write the function as $y=2\log_e\left(x+1\right)-3$`y`=2`l``o``g``e`(`x`+1)−3:

$R:$R: $y$y |
$=$= | $2\log_e\left(x+1\right)-3$2loge(x+1)−3 |

$\therefore R^{-1}:$∴R−1: $x$x |
$=$= | $2\log_e\left(y+1\right)-3$2loge(y+1)−3 |

$\frac{x+3}{2}$x+32 |
$=$= | $\log_e\left(y+1\right)$loge(y+1) |

$e^{\frac{x+3}{2}}$ex+32 |
$=$= | $y+1$y+1 |

$y$y |
$=$= | $e^{\frac{x+3}{2}}-1$ex+32−1 |

$f^{-1}\left(x\right)$f−1(x) |
$=$= | $e^{\sqrt{x+3}}-1$e√x+3−1 |

Consider the function $f\left(x\right)=\log_5\left(x-3\right)$`f`(`x`)=`l``o``g`5(`x`−3) for all $x>3$`x`>3.

By replacing $f\left(x\right)$

`f`(`x`) with $y$`y`, find the inverse function.Leave your answer in terms of $x$

`x`and $y$`y`.State the domain of the inverse function using interval notation.

State the range of the inverse function using interval notation.

Rewrite $y=2\log_ex-3$`y`=2`l``o``g``e``x`−3 with $x$`x` as the subject of the equation.

Consider the function $f\left(x\right)=6\log_e5x-3$`f`(`x`)=6`l``o``g``e`5`x`−3 for $x>0$`x`>0.

By replacing $f\left(x\right)$

`f`(`x`) with $y$`y`, find the inverse function.Leave your answer in terms of $x$

`x`and $y$`y`.State the domain of the inverse function using interval notation.

State the range of the inverse function using interval notation.

For exponential models, express as a logarithm the solution to ab^(ct) = d wherea, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology.