Every equation implies at least two other equation. For example: The equation 2 3= implies 6 2 = 3 and 6 3= =6 6 3=2

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1 Roert C. Maell, 0 INTRODUCTION TO LOGARITHMS Every equation implies at least two other equation. For eample: The equation = 6 implies 6 = and 6 =. =6 6 = 6 = The information in each equation is equivalent it is simply a matter of point of view. Note: For the record, in the multiplication equation = 6 : is the multiplier, is the multiplicand, and 6 is called the product. In the division equation 6 = : 6 is the dividend, is the divisor, and is the quotient. I find the change in the naming of the parts of the equation due to a shift in point of view to e awkward. I will try to avoid that it of awkwardness (awkwardity?) as we proceed. In a similar way consider the eponential equation equations: 8= and 8= =8 = 8. It, too, implies two other 8= 8= In the equation that we will consider our starting point, = 8 : the is called the ase, the is the eponent (or power), and we will call 8 the argument. Notice that each equation is solved for each part of our reference equation: the eponential is solved for the argument; the radical equation is solved for the ase; and the third equation, called a arithmic equation, is solved for the eponent. Note: What an equation is solved for indicates the nature of the question eing asked in the equation. In a division equation we are asking, How many times does the divisor go into the dividend? In the eponential equation the issue is, What do you get when you raise the ase to the eponent? In the radical equation we are asking, What ase do I have to raise to the eponent indicated in the radical Roert C. Maell, 0 Page of 7

2 Roert C. Maell, 0 to get the argument? In arithmic equations we ask, To what power must I raise the ase to get the argument? The arithmic equation 8= is read as The arithm-ase- of 8 is., or more informally Log-ase-two of 8 is. It is asking To what power must I raise in order to get the 8? Let s dump these puppies on the floor and see how they ehave. I will list the arithmic equation with its corresponding eponential equation a) 6= corresponds to ) 7= corresponds to c) = corresponds to = 6 = 7 = d) = corresponds to = e) = corresponds to f) 9= corresponds to = corresponds to g) 7= corresponds to h) 9 Roert C. Maell, 0 Page of 7 = = 9 = 7 i) = corresponds to = Eamples a,, and c highlight what we have already discussed. Logarithm are looking for eponents. Eamples d, e and f are there to highlight that a negative arithm indicates a reciprocal relationship etween the ase and the argument. Eample g is there to remind you that a fractional arithm indicates the argument is a root of the ase. Note: Keep in mind that a inde on a radical can also e shown as an eponent y using its reciprocal = =. Note: If you go ack to where we started the multiplication equation implied two division equations. Think aout it, the multiplication equation is usually considered the simpler, more familiar, equation. Consider that an addition equation implies two sutraction equations, and addition is usually considered the starting place. If = 9

3 Roert C. Maell, 0 you think of radical equations as simply reciprocal power equations, then a arithmic equation implies two eponential equations. I wonder what that means.. Some other oservations on arithms: = This is a simple recognition that =. In general, =. = 0 This follows from the eponential property that ase, = 0 0 =. Regardless of We are going to play in safe, nice, (relatively) pleasant pastures. We are going to keep our values simple. Thus o For our purposes. the ase of a has got to e a positive numer. That means neither 0 nor negative values are candidates for arithmic ases. o Also the argument of a arithm must e a positive value. Think of it this way: Is there a real numer eponent you could raise to, in order to get a negative value? The answer is No. (Technically, the answer is Hellno!, ut pulishers get real ecited aout things like that.) A real numer eponent will never change the polarity of a positive ase. o When you get more eperience with arithms, comple numers, and other aspects of mathematics, then you can get fancy with your arithm. Right now we will keep it simple. Just keep in mind, there is more to this story. PROPERTIES OF LOGARITHMS All properties of arithms are ased on corresponding properties of eponents. Property of eponents = k l k l + Property of arithms ( y) = ( ) + ( y) Eample: Find ( 8). We know 8= =, ut we could also look at it as: ( ) ( ) ( ) 8 = + 8 = + = Property of eponents k l = y = Property of arithms ( ) ( y) k l 8 Eample: Find. We know 8 =, ut we could also look at it as: Roert C. Maell, 0 Page of 7

4 Roert C. Maell, 0 Property of eponents 8 = ( 8) ( ) = = k ( ) l kl = y Property of arithms ( ) = y ( ) Eample: Find ( ). We could calculate = = 0, and then find out how many s we would have to multiply to get 0. You would find you must multiply 0 of them together to get 0. But we could look at it in this way: ( ) ( ) = = = 0 6 Eample with teeth: What is 8? We need to keep in mind that 8= = = 6, and that is the square root of. Then we proceed y any one of three options:. ( ) 8= = + = + 0.=. 6. 8= = 6 = 0.=.. 8= ( 6) = 6 = 6= ( ) =. Now you try it 6 What is? Use = 6= = 0 for your solutions. () = ( 6) = ( ) + ( ) = ( ) + ( )=( ) () = (6/) = ( ) ( )= ( ) ( )=( ) () = ( ) = ( ) ( ) = ( ) ( )=( ) Roert C. Maell, 0 Page of 7

5 Roert C. Maell, 0 Eample with teeth, claws and ad reath: Find Take it one step at a time: = = ( ( ) ( 69 )) = 8 69 = ( 8 ) = ( 6) = ( ( ) ( )). First, do not panic. Now you try it: Determine to If you do the steps right, you ll work your way Property of eponents = (Yes, that looks stupid. Think simple.) k k Property of arithms = and = This means that s and eponentiation are inverse functions of each other. Useful fact = = y y + In any event, what if you are told that ( ) =? Could you deduce what has to e? There are several ways of looking at this equation. The simple approach is est. Since the ase is the same numer that is eing eponentiated, we know + ( ) = +, so what we have is + =. We conclude that =. Note: The constant e (e = ) is the ase of natural s. Natural s are symolized ln. Using e as a ase may seem to have its drawacks. Can you think of any convenient powers of e? Aside from 0, neither can I. Regardless, e is woven into the faric of the universe and is defined y many natural processes having nothing to do with people, let alone mathematicians. Fortunately, the arithmetic is still the same. The rules still hold. So ( y) ln = ln+ lny, etc. Roert C. Maell, 0 Page of 7

6 Roert C. Maell, 0 Now, what if you are told that ( ) that + = 7, so we conclude =. ln e + = 7. Since e is the ase of natural s we see My calculator only does common ase-0 arithms and natural arithms. If I want a ase of 0, I have to do a little it of work. See if you can follow this flow of thought, as I try to find the value of (0) :. I know (0) is some numer. I will call it. So.I write:. (0) =.. There is an eponential equation for every equation. So I write:. = 0. Since I have ln on my calculator I will take ln of oth sides of the equation in step. 6. ln = ln0 7. I can use property and my algera to solve for 8. ln = ln0 ln = ln0 = ln 9. My calculator can handle that last it, so... ln0 0. ( 0) =.07 ln This is the way that a change of ase is usually handled. It can e summarized as: Property of arithms a= a Property Simple algera ln0 The assumption for Property is that I have access to s-ase-, ut what I want is a -ase-a. What you have access to on a standard scientific calculator are natural s and common (ase-0) s. So the useful part of Property is: ln Useful Part of Property a= = lna a Thus you can determine s of whatever ase you desire. For eample, determine 7 ( ), ( 0 ), and ( ) 7 0. ln0 ( 0) = = 0 ln ln0.789( 0 ) = ln.789 Roert C. Maell, 0 Page 6 of ( ) ln( 7) ln ( ) = =

7 Roert C. Maell, 0 Logarithms show up in the darnedest places: chemistry, physics, economics, accounting, and medicine. Often they are manifested in the form of half-lives, doulingtimes or continuous compounding of interest prolems. Eample: A patient takes an anticoagulant the night efore he is to undergo surgery. He cannot undergo the surgery if his lood level of the anticoagulant is over 0.0 mg/dl. The dose he took at :00 p.m. gave him an (immediate) initial lood level of 0.9 mg/dl. The anticoagulant has a ioic half-life of. hours. Will he e ale to undergo is surgery at the scheduled time of 7:00 am? How soon after his :00 p.m. dose of anticoagulant will his lood anticoagulant level reach 0.00 mg/dl? Solution: The half-life indicates the time it will take for one-half the amount of the material to e inactive or gone away in some fashion. In this case, after. hours half of the drug that is there goes away. We ask how many. hour increments are there from pm to 7 am. So we divide 8 hours y. to get,.. So the amount that remains in the pilot s lood is:. 0.9 = 0.06 mg/dl So the answer to the first question is, No. He cannot the surgery at 7:00 a.m. The second prolem can e written in the form: 0.9 = 0.00 This can e solved y taking s of oth sides of the equation: ln 0.9 = ln( 0.00) ln( 0.9) + ln = ln( 0.00) ln( 0.9) + ln = ln( 0.00) ln( 0.00) ln( 0.9) = = 8.79 ln So the answer is: 8.79 half-lives, or (8.79)(.) = 0.8 hours. He will have to wait at least 0.8 hours efore his anticoagulant level is 0.00 mg/dl. Roert C. Maell, 0 Page 7 of 7