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Working with Logarithms

A logarithm is an exponent.

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As this example shows, 3 is the exponent to which the base 2 must be raised to create the answer of 8,
or  23 = 8.  In general terms:

logs
(where x > 0 and b is a positive constant not equal to 1)
BASE 10:
Logarithms with base 10 are called common logarithms.
When the base is not indicated, base 10 is implied.
The log key on the graphing calculator will calculate the
common (or base 10) logarithm.
2nd log will calculate the antilogarithm or 10x
logpic1
log1

 

OTHER BASES:
To enter a logarithm with a different base on the graphing calculator, use the Change of Base Formula:

log7 arrowlog64pic1


If your calculator has the logBASE operation template, you can enter the values as they appear in the expression.
 Go to: MATH → arrow down to A:logBASE(
Or hit MATH, ALPHA (key), MATH to get to the A: option.
log2
alpha


You can also hit ALPHA (key), WINDOW
and choose the fifth option on the menu, logBASE(.

 
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BASE e:

Logarithms with base e are called natural logarithms.
Natural logarithms are denoted by ln.
On the graphing calculator, the base e logarithm is the ln key. 

monster logh2
e log98 2.71828183
log4

All three are the same.

 

       When working with logarithms on your graphing calculator,
you must remember the "Change of Base Formula":

log16
Remember, the notation:
 log x is with respect to base 10
ln x is with respect to base e


Examples:

1.  Evaluate:   log94

lp
 
2.  Evaluate:  ln 2;  ln 1;  ln e

l2
3.  Evaluate:   l9

l10
4.  Sketch the graph of  l18

If you have the logBASE function, it can be used to enter the function (seen in Y1 below).
If not, use the Change of Base formula (see in Y2 below).
log3b

l2
5.  Use a graph to support the conclusion that the inverse of  l19 is l20.

A function composed with its inverse creates the identity line y = x.  Showing that the composition of these two functions is the identity line shows that these functions are inverses of one another.  Note the bubble animation set for Y4 to show that the linear lines in Y3 and Y4 are equivalent.
You are showing that the composition of Y1(Y2) = x (the Identity Function).

l3
6.  Solve graphically:    l21

 

Method 2:  This problem can also be solved by setting the equation equal to 0 and finding the x-intercepts, or zeros (2nd CALC, #2), of the function.
l99
l4

Method 1: Graph each side separately.
The window must be adjusted to show a sufficient amount of the x-axis to locate the intersection point (2nd CALC, #5).  The answer is x = 32.
l4
7.  Using your graphing calculator, determine which of the following statements are true.
l27

 

a.) T  b.) F  c.) F  d.) T  e.) F  f.)  T
Solution for part a:  (TRUE)
Place the left side of the equation in Y1 and the right side in Y2.  Turn on the "bubble animation" to the left of the Y2.  This will allow you to see the bubble floating over the graph when the equation is true.  Parts b - f are solved in a similar manner. 

l5


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