The Log Function |

**LOG FUNCTIONS**

**LOGS ARE EXPONENTS!! so they behave like exponents!!**

The log function is the inverse of the exponential functionif y = a then ^{ x} x = log _{a} y (log base a of y) |

*log _{a} y* is pronounced

When we use an *exponential function*, we **input exponents** and **haul out numbers**.

For example if *f *(*x*)* = *5* ^{ x}* + 1

With the log function, we do the opposite.

We

So

Notice that the "

**LOGS ARE EXPONENTS!!**

We learned in the exponential function, that all exponential bases are positive or > 0.

The same is true for logs. So if *f *(*x*)*= log _{ }*3

The **domain** of the log function therefore is all **positive Real numbers**.

* f *(*x*)**= a log _{ c} [b (x – h) ] + k** has a

** note: y = ln x **means

For both these bases (*e* and *10*) we don't indicate the base. We assume it.

There are two sets of log functions depending on whether **base c is greater than or less than 1**. **All log bases are positive**. They can be** fractions between 0 and 1** or **greater than one**.

That is, **0 < c < 1** produces one set of log functions

while **c > 1** yields another set of log functions.

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**Log Graphs**

The graph displays the function *y = log _{ c} x*

The curve is always **increasing** if **c > 1**

The curve is always **decreasing** if **0 < c < 1**.

Notice that the domain and range are the same,

**domain**: *x* > 0, **range**: *R*

Both curves move right since ** b** is positive.

Here is how the values of *a, b *and* c *affect the graph:

c > 1 |
if a > 0, curve moves up |
if a < 0, curve moves down |

0 < c < 1 |
if a > 0, curve moves down |
if a < 0, curve moves up |

all cases | b > 0, curve moves right |
if b < 0, curve moves left |

domain: | if b > 0 |
if b < 0 |

range: R |
zero: always one and only one |

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.Hints on graphing log functions.

1) put the rule in standard form *f *(*x*)* = a log _{ c} *[

2) look at *c* to determine which class of log function it is -- (is *0 < c* < 1 or c > 1?)

3) look at *a *and* b* to decide if the curve moves up, down, left or right from the asymptote.

4) put in the vertical asymptote *x = h*, and the zero then graph the curve.

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let's solve *0 = a log _{ c} *[

So, if *f *(*x*)* = 3 log _{ 2} *(

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Let's graph *f *(*x*)* = 5 log *(2*x* + 8) – 6.

First we'll put the rule in standard form by factoring out the 2 in (2*x* + 8).

So *f *(*x*)* = 5 log* 2 (*x *+ 4) – 6.

The vertical asymptote is *x* = – 4

The *y-intercept *= – 1.48 (set *x* = 0)

The *zero* = 3.92

Since *c* = 10 > 1*, a *and* b* > 0, the curve is increasing, and moves right from the asymptote. .

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**Practice**

1/ If a log function rule is of the form *f* (*x*)* = log _{ c} *(

a) points P ( – 2, 0) and Q (3, 1) are on the curve.

b) points P (4, – 1) and Q (6, – 2) are on the curve.

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2/ *f* (*x*)* = *–* log _{ 2} *(3

a) put the rule in standard form.

b) what is the equation of the vertical asymptote?

c) is the curve always increasing or always decreasing?

d) does the curve move right or left from the asymptote?

e) find the zero and the *y*-intercept if any.

f) graph the curve.

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3/ For each of these log functions put the rule in standard form and state:

a) the equation of the asymptote | b) increasing or decreasing? | |

c) moves left or right | d) the zero | |

f (x) = – 2 log_{ 5} (3 – 6x) – 1 |
g (x) = 7 log_{ 1 / 3} (2x + 1) |
h (x) = 6 log_{ 4} (1 – 3x) + 2 |

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**Solutions**

1/ a) Substituting P ( – 2, 0) gives us *0 = log _{ c} *(–2 –

Now we find *c* by substituting Q (3, 1)

1 = *log _{ c} *(3 – (–3)) which becomes

*f *(*x*) =* log*_{ 6} (*x* + 3)

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b)Substituting P (4, – 1) becomes – 1* = log _{ c}* (4 –

substituting Q (6, – 2) becomes – 2 =* log _{ c}* (6 –

Since *c*^{ – 2} = (*c*^{ – 1})^{ 2}* *, (4 – *h*)^{ 2} = 6 – *h* becomes 16 – 8*h + h*^{ 2} = 6 – *h*.

When we solve the quadratic, we get that *h* = 5 or *h* = 2

Since *c*^{ –1} = 4 – *h*, and *c > 0* we must reject the solution that *h* = 5

If *h* = 2, *c*^{ – 1} = 4* *–* h* = 2* *so* c* = ½

*f* (*x*) = *log _{ }*1 / 2 (

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2/ *f* (*x*)* = *–* log*_{ 2} (3*x* + 3) – 5

a) *f* (*x*)* = *–* log*_{ 2} 3(*x* + 1) – 5

b) the vertical asymptote is *x* = – 1.

c) since *a < 0 *, always decreasing.

d) since *b > 0* the curve moves right.

e) the *zero* = – 0.99 * *(solution below)* and the *y-intercept* = 6.58

f)* finding zero: *0 = *–* log _{ }*2 3(

applying 3rd rule of logs gives us 5 = *log*_{ 2} (3*x* + 3)^{ –1}* *so

since 2^{ 5} = 32, the equation becomes 32[3(*x* + 1) = 1, so *x =* – 1 + (1 / 96) – 0.99

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3/ For each of these log functions put the rule in standard form and state:

f (x) = – 2 log_{ 5} (3 – 6x) – 1 |
g (x) = 7 log_{ 1 / 3} (2x + 1) |
h (x) = 6 log_{ 4} (1 – 3x) + 2 |

f (x) = – 2 log_{ 5} – 6(x – ½) – 1 |
g (x) = 7 log_{ 1 / 3} 2(x + ½) |
h (x) = 6 log_{ 4} – 3(x – 1/3) + 2 |

a) asymp is x = ½ |
a) asymp is x = –½ |
a) asymp is x = 1/3 |

b) decreasing | b) decreasing | b) increasing |

c) moves left | c) moves right | c) moves left |

d) zero is – 0.4255 |
d) zero is 0 | d) zero is – 0.12333 |

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