The Log Function

LOG FUNCTIONS

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

 The log function is the inverse of the exponential function if y = a x then x = log a y (log base a of y)

log a y is pronounced log base a of y.

When we use an exponential function, we input exponents and haul out numbers.
For example if f (x) = 5 x + 1 , f (0) = 5 and f (2) = 5 3 = 125
With the log function, we do the opposite.
We input positive numbers for x, and haul out exponents for f (x).
So log 7 49 = 2 is of the form log c x = y where c = 7 .
Notice that the "log" is equal to 2 -- the exponent -- because

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 x, we can only substitute positive values for x, since there are no powers of a positive base that will give us negative numbers.

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

f (x)= a log c [b (x – h) ] + k has a vertical asymptote at x = h.

note: y = ln x means y = log e x. ln x is pronounced "lawn" x (just like your front lawn). These are called natural logs (actually ln indicates logarithmes naturels from French)
y = log x means y = log 10 x. Logs base 10 are called common logs.

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 [b (x – h) ] + k

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 [b (x – h) ] + k for x to find an expression for the zero.

So, if f (x) = 3 log 2 (x – 1) + 5, the zero would be

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

First we'll put the rule in standard form by factoring out the 2 in (2x + 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 (x – h), find c and h given:

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 (3x + 3) – 5

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 – h) becomes 1 = – 2 h so h = – 3

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

1 = log c (3 – (–3)) which becomes c1 = 6 in exponential form, so c = 6.

f (x) = log 6 (x + 3)

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b)Substituting P (4, – 1) becomes – 1 = log c (4 – h) becomes c – 1 = 4 – h

substituting Q (6, – 2) becomes – 2 = log c (6 – h) becomes c – 2 = 6 – h

Since c – 2 = (c – 1) 2 , (4 – h) 2 = 6 – h becomes 16 – 8h + 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 (x – 2)

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2/ f (x) = log 2 (3x + 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(x + 1) – 5

applying 3rd rule of logs gives us 5 = log 2 (3x + 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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