log notes explanation day 1
TRANSCRIPT
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Introduction To
Logarithms
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Logarithms were originallydeveloped to simplify complex
arithmetic calculations.
They were designed to transformmultiplicative processes
into additive ones.
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If at rst this seems like no big deal,then try multiplying
2,2!,!"#,#$2 and ,!"%,2!,!"#.
&ithout a calculator '
(learly, it is a lot easier to addthese two numbers.
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Indeed, they would be obsolete except for onevery important property of logarithms.
It is calledthe power property and we
will learn about it in another lesson.
*or now we need only to observe thatit is an extremely important partof solving exponential e+uations.
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ur rst -ob is to
try to makesome sense of
logarithms.
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ur rst +uestion
then must be
&hat is a logarithm /
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f course logarithms have
a precise mathematicaldenition -ust like all terms in
mathematics. )o let0s
start with that.
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1enition ofLogarithm
)uppose b3 and b4$,
there is a number 5p0such that
logb n = p if and only if b p = n
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6ow amathematician
understands exactlywhat that means.
7ut, many a
student is leftscratching theirhead.
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The rst, and perhaps
the most important step,in understandinglogarithms is to reali8e
that they always relateback to exponential
e+uations.
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9ou must be able toconvert an exponential
e+uation into logarithmicform and vice versa.
)o let0s get a lot of practice withthis '
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:xample $
)olution log2 8 = 3
&e read this as ;thelog base 2 of < is e+ual
to ;.
3Write 2 8 in logarithmic form.=
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:xample
$a
Write 42 = 16 in logarithmic form.
)olution log4 16 = 2
=ead as >the logbase ! of $% is
e+ual to 2;.
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:xample $b
)olution
Write 2− 3 =
1
8 in logarithmic form.
log21
8= − 3
1
Read as: "the log base 2 of is equal to -3".8
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kay, so now it0s timefor you to try some on
your own.
1. Write 72 = 49 in logarithmic form.
7Solution: log ! 2=
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log5 1= 0)olution
2. Write 50
= 1 in logarithmic form.
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3. Write 10− 2 =
1
100
in logarithmic form.
)olution log101
100 = − 2
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)olution log16 4 = 12
4. Finally, write 161
2= 4
in logarithmic form.
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:xample 2
Write log21
8= − 3 in exponential form.
)olution 2− 3
=
1
8
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kay, now you try thesenext three.
1. Write log10 100 = 2 in exponential form.
3. Write log27 3 =1
3 in exponential form.
2. Write log51
125
= − 3 in exponential form.
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1. Write log10 100 = 2 in exponential form.
)olution 102 = 100
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2. Write log51
125= − 3 in exponential form.
)olution 31
"
12"
−
=
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3. Write log27 3 =1
3 in exponential form.
)olution 271
3 = 3
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&e now know that alogarithm is perhaps best
understood
as beingclosely related to an
exponential e+uation.
In fact, whenever we getstuckin the problems that follow
we will return tothis one simple insight.
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&hen working with logarithms,if ever you get >stuck;, try
rewriting the problem inexponential form.
(onversely, when working
with exponential expressions,if ever you get >stuck;, try
rewriting the problemin logarithmic form.
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)olutionLet0s rewrite the
problem inexponential form.
62
= x
&e0re nished '
#Sol$e for %: log 2 x =
&%am'le 1
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)olution
5 y = 1
25
=ewrite the problemin exponential form.
Since 1
25= 5− 2
5 y
= 5− 2
y = −2
1Sol$e for (: log
2 y=
&%am'le 2
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:xample
Evaluate log3 27.
Try setting this up like this
)olution
log3 27 = y 6ow rewrite in exponentialform.3 y = 27
3 y = 33
y = 3
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These next twoproblems tend to besome of the trickiest
to evaluate.
?ctually, they are
merely identities andthe use of our simple
rulewill show this.
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:xample !
Evaluate: log7 72
)olution
6ow take it out of the logarithmic
formand write it in exponential form.
log7 72
= y*irst, we write the problem with a variable.
7 y
= 72
y = 2
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*inally, we want to take alook at the @roperty of
:+uality for Logarithmic
*unctions.Suppose b > 0 and b ≠ 1.Then logb x1 = logb x 2 if and only if x1 = x 2
7asically, with logarithmic functions,if the bases match on both sides of the
e+ual sign , then simply set thearguments e+ual.
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:xample $
Solve: log3 (4 x +10)= log3 ( x +1)
)olution)ince the bases are both 50 wesimply set the arguments e+ual.
4 x +10 = x +13 x +10 = 13 x = − 9 x = − 3
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:xample 2
Solve: log8 ( x2 −14) = log8 (5 x)
)olution)ince the bases are both 5
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:xample 2continued
Solve: log8 ( x2 −14) = log8 (5 x)
)olution x = 7 or x = −2It appears that we have 2 solutions
here.
If we take a closer look at thedenition of a logarithm however, wewill see that not only must we use
positive bases, but also we see thatthe arguments must be positive as
well. Therefore A2 is not a solution.Let0s end this lesson b takin a
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ur nal concern then is todetermine why logarithms
like the one below are
undened.
(an anyone giveus an
explanation /
2log ) 8*−
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ne easy explanation is to simplyrewrite this logarithm in exponential
form.
&e0ll then see why a negative valueis not permitted.
*irst, we write the problem with a variable.
2 y = − 8 6ow take it out of the logarithmic
formand write it in exponential form.&hat power of 2 would gives us A< /
23
= 8 and 2− 3
=
1
8
Bence expressions of this type are
undened.
2log ) 8* undefined W+,− =
2log ) 8* y− =
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That concludes our introductionto logarithms. In the lessons to
follow we will learn some important
properties of logarithms.
ne of these properties will giveus a very important tool
whichwe need to solve exponential
e+uations. Cntil then let0spractice with the basic themes
of this lesson.