Post on 05-Jan-2016
description
INTERESTING QUESTIONS &“UNDOING” EXPONENTS: LOGS
Hey, hey logarithm…When we get the blues
Recap
Last week we looked at RATIONAL exponents and saw that
A square root is the same as an exponent of ½ A square root is the same as an exponent of ½
A cubed root is the exponent 1/3 A cubed root is the exponent 1/3
To evaluate powers with rational exponents, we “rip the exponent apart”. To evaluate powers with rational exponents, we “rip the exponent apart”.
We also saw that radioactive materials will decay in an exponential fashion (half-life)We also saw that radioactive materials will decay in an exponential fashion (half-life)
We also saw that compound interest can be modeled using exponential equationsWe also saw that compound interest can be modeled using exponential equations
InterestInterest
In general, the compound interest is nt
n
rPA
1
Where A is the amount in the account at time, tP is the principle (initial) amount r is the decimal value of the interest rate n is how many times per year the interest is compounded.
Look for terms like: daily (n =365), semi-annually (n = 2), weekly (n = 52) and monthly (n =12)
Look for terms like: daily (n =365), semi-annually (n = 2), weekly (n = 52) and monthly (n =12)
InterestInterest
Ex 1. A credit card charges 24.2% interest per year compounded monthly. There are $900 worth of purchases made on the card. Calculate the amount owing after 18 months. (Assume that no payments were made.)
)(12
12
242.01900
t
A
)5.1(12020166667.1900A
21.1289$A
A new type of question?
Ex 2. A bank account earns interest compounded monthly at an annual rate at 4.2%. Initially the investment was $400. When does it double in value?
So this questions seems to be like all the others…
t
y12
12
042.01400
t120035.1400800
t120035.12
And now we get
Solving for the Exponent
t120035.12We’re totally stuck! We presently have no way of solving for an exponent unless we can get common bases.
We’re totally stuck! We presently have no way of solving for an exponent unless we can get common bases.
So mathematicians invented logarithms.
823 can also be written as
38log2
basebase
exponentexponent
argumentargument
argumentargument
basebase
exponentexponent
We read this “log to base 2 of 8 equals 3.We read this “log to base 2 of 8 equals 3.
Going from one form to another
932
Write the following in logarithmic form.
81643
25
1125 3
2
130
1000103
63621
29log3
2
16log36
4
38log16
3
2
25
1log125
01log3
31000log When the base is a ‘10’ we do not need to write it. This is because base 10 is what most calculators deal with.
When the base is a ‘10’ we do not need to write it. This is because base 10 is what most calculators deal with.
Going from one form to another
Write the following in exponential form.
5.09log81
3
1
5
1log125
3
24log8
2
150log50
29
4log
23
201.0log
981 5.0
5
1125 3
1
4832
505021
9
4
2
32
01.010 2
cab log
abc
Evaluating Logs
Evaluate: a) 32log2 This asks “2 to the what gives 32?”
or 322 x We know this is 5.
532log2
b) 64log4 This asks “4 to the what gives 64?”
or 644 x
We know this is 3.364log4
c) 32log41 This asks “1/4 to the what
gives 32?”32
4
1
x
We can get common bases
52 22 x
5.2
52
x
x
By changing forms we can evaluate log expressions.
d) log 100 This asks “10 to the what gives 100”?
We know this is 2.
5.232log41
Solving log equations: common bases
To continue getting used to logs, we’ll look at these questions.
Solve for x.
3log2 x
x 32
We’re stuck in log form so go to exponential form
We can solve this: x =1/8
b) x9log3
We’re stuck in log form so go to exponential form
a)
22
22
1
33
33
93
x
x
x
4
22
x
x
c) 5
416log x
We’re stuck in log form so go to exponential form
165
4
x To solve for the base, we can undo the exponent by raising both sides to the 5/4.
4
54
5
5
4
16
x
5
4
1
4
5
16
16
x
x 32
2 5
x
x
d) x3.1log
We’re stuck in log form so go to exponential form
But wait… the base here is
10
Since the calculator uses base 10 I just type this in and get…
log1.3=0.114
Solving log equations: common bases
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
5010 x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
x 27log32log 32a.
x 27log5 3
y27log3
273 y
2
3
33
33
2
3
2
13
y
y
y
2
72
3
2
102
35
x
x
x
x272 log16log3
1
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
5010 x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
b.
2log2 x
4
22
x
x
100loglog2 x
x272 log16log3
1
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
5010 x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
c.
38log2
x632 log27
1log8log
327
1log3
x
x
6
6
log0
log)3(3
1
60
x
x
x272 log16log3
1
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
5010 x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
d.
14
1log4
481log3
2
21
x
x
x25
134 log25log81log4
1log
225log5
1
x
x
2
2
log1
log)2(41
x272 log16log3
1
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
5010 x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
e.
5010 x
Since I can’t get common bases, I’m stuck in exponential form. So I go to log form.
x50log10
x50log My calculator can find this.
My calculator can find this.
70.1x
x272 log16log3
1
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
f.
5010 x
I’m stuck in log form. So I go to exponential form.
9
43
2
x
3
2
9
4log x
To solve for the base, I undo the exponent. I raise both sides to the -3/2
2
32
3
3
2
9
4
x
2
3
4
9
x
The negative in the exponent means I “flip the base”
3
2
1
4
9
x 3
2
3
x 8
27x
x272 log16log3
1
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
x272 log16log3
1
g.
5010 x
x272 log16log3
1
x27log)4(3
1
416log2
x27log3
4
I’m stuck in log form. So I go to exponential form.
I’m stuck in log form. So I go to exponential form.
x3
4
27 Rip the exponent apartRip the exponent apart
x
4
3
1
27 x43 x81
a.
b.
c.
d.
e.
f.
g.
h.
Try some…
1) Solve each equation
x 27log32log 32
100loglog2 x
3
2
9
4log x
x25
134 log25log81log4
1log
)4(log5log 22 x
x632 log27
1log8log
x272 log16log3
1
h.
5010 x
)4(log5log 22 x
I don’t know what is.
5log2
But I do notice that there’s a common base on both sides of the equation.
Since the bases are equal, the ARGUMENTS must be equal.
9
)4(5
x
x
So What?We can now write an equation where we solve for the exponent but how do we evaluate the exponent if we do not recognize the log or if we cannot get common bases?
All of these started with the equation
t120035.12 Are we any closer?
Well we can write it in a log form
t122log 0035.1 But that’s as far as we can get.
Notice that theis equal to an exponent. It must also be an exponent.
2log 0035.1
How do the laws of exponents relate to logs?
The Laws of Logs
Remember the laws of exponents:
When multiplying powers with the same base, we keep the base and add the exponents.
When multiplying powers with the same base, we keep the base and add the exponents.
cbcb aaa
xab yac
cbayx ))((
Let
Go to log form cbxya log
yxxy aaa logloglog When adding logs with the same base, we keep the log and base and multiply the arguments.
When adding logs with the same base, we keep the log and base and multiply the arguments.
If then xb alog xab If then yc alog yac
The Laws of Logs
Remember the laws of exponents: cb
c
b
aa
a
xab yac
cbay
x
Let
Go to log form
cby
xa
log
When dividing powers with the same base, we keep the base and subtract the exponents.
When dividing powers with the same base, we keep the base and subtract the exponents.
If then xb alog xab If then yc alog yac
yxy
xaaa logloglog
When subtracting logs with the same base, we keep the log and base and divide the arguments.
When subtracting logs with the same base, we keep the log and base and divide the arguments.
Laws of Exponents
Practice with the first two laws. Solve for x.
25log2log8log4log.2# 5 xxx
22log)84(log xx
22
32log
x
216log xI’m stuck in log form. So I go to exponential form.
I’m stuck in log form. So I go to exponential form.
4
162
x
x
x233 log7log189log.1#
x23 log7
189log
x23 log27log
x2log3
8
23
x
x
The 3rd Law of Logs
When we have a power of a power, we keep the base and multiply the exponents.
When we have a power of a power, we keep the base and multiply the exponents. Le
t ant
If then at nlog ant
Raise both sides to the exponent “b”
bbt an btb an Go to log form
bn abt log bnn aab loglog
The “down in front” rule!
Ex. Evaluate
27log3
21
3 27log
27log2
13
2
3
And finally…
Are we any closer to solving the original question?
t120035.12 Let’s take the log of both sides…
t120035.1log2log Now the “down in front” rule
0035.1log122log t Divide by 12log1.0035
t0035.1log12
2logAnd my calculator can do this
53.16t
A shortcut to the calculator rule
So we have seen that t120035.12 can be written as
t120035.1log
2log
So we do not need to take the log of both sides. We can go to log form
t122log 0035.1 And then write t120035.1log
2log
Remember that the base is on the bottom!
Lots o’ Logs
Solve for x.
1003 9 x Since I can’t get common bases, I’m stuck in exponential form. So I go to log form.
9100log3 x
93log
100logx
919.4 x 81.4x
Lots o’ Logs
Solve for x.
8047 x
x74log
80log
Since I can’t get common bases, I’m stuck in exponential form. So I go to log form.
x780log4
84.3
716.3
716.3
x
x
x