Finding Equations Indirectly

Here is a set of ordered pairs, ( x, y ) , for which we want to find a generating function.

x 1 2 3 4 5 6 7 8

z = log ( y )

y

6 12 24 48 96 192 384 768

It’s really hard to graph these ordered pairs because, well, 768 is just so much bigger than 384, never mind how much bigger than 6 it is (it’s 762 bigger in case you’re wondering…). Let’s see how logs can help with this. First fill in the column for z = log ( y ) . You should probably just use the actual values, which is a sort of silly way to fill in a column, but…do it. If you know how, create a scatter plot of the ordered pairs ( x, z ) . You should notice that this is a linear function! If you don’t know how, here’s a scatter plot I made in GeoGebra—along with a graph of the original data. Original

Cool Log Things

Logged

1

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Pick any two ordered pairs ( x, z ) and find the slope of the line passing through them. Try it for another two points and make sure you get the same answer—this works because I made up a perfect function for this process. If the data isn’t quite so perfect you might just get something close, rather than exactly the same.

Write the equation of the line containing the ordered pairs ( x, z ) using point-slope form.

Now replace z with log ( y ) and solve for y.

Do the original ordered pairs ( x, y ) fall on the function you just found?

Why does this work? Since logs and exponentials are inverses of each other, if you take the log of a set of data that came from an exponential function, you should end up with a linear set of data…then we can find that equation, make a replacement, and find the original equation. Isn’t that neat? Are you not entertained?

Cool Log Things

2

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Determining How Many Digits a Number Has A common weird question on math contests and some standardized tests (hard standardized tests) is to figure out how many digits a number will have. We can use logs to answer that question. First of all, what is each of the following logs equivalent to? log (1) log (10 ) log (100 ) log (1000 )

log (10000 )

Now, logs increase. You can see that in the table, but you can just feel it, right? It makes sense. So how is this helpful? Well, until you get to x = 10 , log ( x ) will be less than 1. Similarly until you get to x = 100 , log ( x ) will be less than 2. Now fill in this table. x

1 ≤ x < 10 10 ≤ x < 100

log ( x )

Digits of x

0 ≤ log ( x ) < 1

1

1 ≤ log ( x ) < 2

2

100 ≤ x < 1000 1000 ≤ x < 10000 So, how does the log of a number relate to the number of digits the number has?

Use your calculator to figure out how many digits each of the following numbers has. a. ( 333) ( 555 ) ( 888 ) b. 35 5 7 7 9

( )( )( )

c. 75!

d. 449!

449! is actually the largest factorial you can get the TI-Nspire to give you the log of…after that it’s just an overflow error. Anyway, this is one of my favorite applications of logs. Cool Log Things

3

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Dividing and Multiplying Huge Numbers I’m not totally positive, but I suspect this is really why logs were invented. Say I want to do something like the division problem below. 1056747483332 95834847321 Well, that would be awful right? Kind of…but then someone (John Napier, actually, of Napier’s Bones fame—google that) invented logs. What am I finding if I calculate log (1056747483332 ) ?

With that in mind, how could I rewrite the numerator of that fraction?

Do the same thing for the denominator.

Rewrite the fraction.

Simplify the fraction.

Now confirm that both versions of your fraction are equivalent. Are they? But how did the ancients do this? We used our calculators, back in the day they used to have tables of logs. Literally a book with just lists and lists of the logs of various numbers. There were also well-established methods of approximating the logs of numbers that weren’t in the list but were between numbers that did show up in the book. Anyway, they could have figured out that log (1056747483332 ) ≈ 12.024 and log ( 95834847321) ≈ 10.982 . Armed with this, they would have been able to figure that:

1056747483332 1012.024 ≈ 10.982 = 101.042 95834847321 10 Cool Log Things

4

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But where do you go from there? Well, you use your table of logs to work backwards and figure out that 1.042 is just about the log of 11.0154. Which is pretty close. If your table of logs goes beyond 3 decimal places, you get better accuracy. We essentially pretended that our calculator was a table of logs. Our calculator holds many, many decimal places. Pretty much everyone agrees that division is awful, so turning it into subtraction is a neat trick. We can also use this idea to multiply numbers. I’ll just work this one out for you. I’m going to do the following problem:

( 52466789 )( 6542231) I actually have the option of doing this with logs instead of just straight away. Remember, if I’m using a table of logs, I’m probably (well, definitely, really) losing accuracy, but I’m gaining time, because it’s much easier. Let’s do it. First I’ll find the log of both numbers, and get:

log ( 52466789 ) ≈ 7.7198844859204 and log ( 6542231) ≈ 6.8157258745881 So now the question is just how accurate I want to be. If I go with 8 decimal places for my log approximations, I’d have:

( 52466789 )( 6542231) ≈ (10 7.71988449 )(10 6.81572587 ) = 1014.53561036 All I have to do now is add the exponents and then look that number up in my log table. My calculator gives me the result in scientific notation, which is sort of annoying. But when I asked it to subtract the approximation from the exact value I get 401540. That seems like a big difference. But calculating the percent error, approximation − actual (100 ) , actual my calculator says the error is 0.00000011698% , which is pretty much nothing. This number is so big that half a million either way doesn’t really make a difference.

Cool Log Things

5

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Cool Log Things Solutions Logs are okay, but there are a few really neat things you can do with them. So let’s do them.

Finding Equations Indirectly

Here is a set of ordered pairs, ( x, y ) , for which we want to find a generating function. y z = log ( y ) x 1

6

2

12

3

24

4

48

5

96

6

192

7

384

log ( 6 )

log (12 )

log ( 24 ) log ( 48 ) log ( 96 )

log (192 )

log ( 384 )

log ( 768 ) 8 768 It’s really hard to graph these ordered pairs because, well, 768 is just so much bigger than 384, never mind how much bigger than 6 it is (it’s 762 bigger in case you’re wondering…). Let’s see how logs can help with this. First fill in the column for z = log ( y ) . You should probably just use the actual values, which is a sort of silly way to fill in a column, but…do it. If you know how, create a scatter plot of the ordered pairs ( x, z ) . You should notice that this is a linear function! If you don’t know how, here’s a scatter plot I made in GeoGebra—along with a graph of the original data. Original

Cool Log Things

Logged

6

www.turksmathstuff.com

Pick any two ordered pairs ( x, z ) and find the slope of the line passing through them. Try it for another two points and make sure you get the same answer—this works because I made up a perfect function for this process. If the data isn’t quite so perfect you might just get something close, rather than exactly the same. log (12 ) − log ( 6 ) ⎛ 12 ⎞ = log ⎜ ⎟ = log ( 2 ) ; ⎝ 6⎠ 2 −1 log ( 768 ) − log ( 48 ) 1 ⎛ 768 ⎞ 1 = log ⎜ = log (16 ) = log 161 4 = log ( 2 ) ⎝ 48 ⎟⎠ 4 8−4 4 It doesn’t matter what we use, we’ll get log ( 2 ) since it’s a line.

(

)

Write the equation of the line containing the ordered pairs ( x, z ) using point-slope form. z − log ( 6 ) = log ( 2 ) ( x − 1) Now replace z with log ( y ) and solve for y.

⎛ y⎞ log ( y ) − log ( 6 ) = log ( 2 ) ( x − 1) ⇒ log ⎜ ⎟ = ( x − 1) log ( 2 ) ⎝ 6⎠ y ⎛ y⎞ ⇒ log ⎜ ⎟ = log ( 2 x−1 ) ⇒ = 2 x−1 ⇒ y = 6 ( 2 x−1 ) ⎝ 6⎠ 6 Do the original ordered pairs ( x, y ) fall on the function you just found? Yeah, they definitely go through all of those original ordered pairs.

Why does this work? Since logs and exponentials are inverses of each other, if you take the log of a set of data that came from an exponential function, you should end up with a linear set of data…then we can find that equation, make a replacement, and find the original equation. Isn’t that neat? Are you not entertained?

Cool Log Things

7

www.turksmathstuff.com

Determining How Many Digits a Number Has A common weird question on math contests and some standardized tests (hard standardized tests) is to figure out how many digits a number will have. We can use logs to answer that question. First of all, what is each of the following logs equivalent to? log (1) log (10 ) log (100 ) log (1000 ) 0

1

2

log (10000 )

3

4

Now, logs increase. You can see that in the table, but you can just feel it, right? It makes sense. So how is this helpful? Well, until you get to x = 10 , log ( x ) will be less than 1. Similarly until you get to x = 100 , log ( x ) will be less than 2. Now fill in this table. x

1 ≤ x < 10 10 ≤ x < 100 100 ≤ x < 1000 1000 ≤ x < 10000

log ( x )

Digits of x

0 ≤ log ( x ) < 1

1

1 ≤ log ( x ) < 2

2

2 ≤ log ( x ) < 3

3

3 ≤ log ( x ) < 4

4

So, how does the log of a number relate to the number of digits the number has? When you take the log of a number you usually get an integer and a decimal part. The integer part actually has a name. It’s called the characteristic. (The decimal part is called the mantissa. No one really seems to use either of those names anymore.) Anyway, the number of digits in the number you’re taking the log of is 1 more than its characteristic. Use your calculator to figure out how many digits each of the following numbers has. e. ( 333) ( 555 ) ( 888 ) f. 35 5 7 7 9

( )( )( ) log (( 3 ) ( 5 ) ( 7 )) ≈ 14.884

log (( 333) ( 555 ) ( 888 )) ≈ 8.215 The product has 9 digits.

5

7

9

The product has 15 digits.

g. 75! log ( 75!) ≈ 109.395 The product has 110 digits.

h. 449! log ( 449!) ≈ 997.586 The product has 998 digits.

449! Is actually the largest factorial you can get the TI-Nspire to give you the log of…after that it’s just an overflow error. Anyway, this is one of my favorite applications of logs. Cool Log Things

8

www.turksmathstuff.com

Dividing and Multiplying Huge Numbers I’m not totally positive, but I suspect this is really why logs were invented. Say I want to do something like the division problem below. 1056747483332 95834847321 Well, that would be awful right? Kind of…but then someone (John Napier, actually, of Napier’s Bones fame—google that) invented logs. What am I finding if I calculate log (1056747483332 ) ? I’m finding the exponent to which 10 must be raised in order to get that really long number that I’m not typing again… With that in mind, how could I rewrite the numerator of that fraction? Bah! There’s no avoiding typing it again… 1056747483332 ≈ 1012.024 Do the same thing for the denominator. 95834847321 ≈ 1010.982

Rewrite the fraction. 1056747483332 1012.024 ≈ 10.982 95834847321 10 Simplify the fraction. 1012.024 = 1012.024−10.982 = 101.042 10.982 10 Now confirm that both versions of your fraction are equivalent. Are they? They are roughly equivalent because I did a whole lot of rounding. But how did the ancients do this? We used our calculators, back in the day they used to have tables of logs. Literally a book with just lists and lists of the logs of various numbers. There were also well-established methods of approximating the logs of numbers that weren’t in the list but were between numbers that did show up in the book. Anyway, they could have figured out that log (1056747483332 ) ≈ 12.024 and log ( 95834847321) ≈ 10.982 . Armed with this, they would have been able to figure that:

1056747483332 1012.024 ≈ 10.982 = 101.042 95834847321 10 Cool Log Things

9

www.turksmathstuff.com

But where do you go from there? Well, you use your table of logs to work backwards and figure out that 1.042 is just about the log of 11.0154. Which is pretty close. If your table of logs goes beyond 3 decimal places, you get better accuracy. We essentially pretended that our calculator was a table of logs. Our calculator holds many, many decimal places. Pretty much everyone agrees that division is awful, so turning it into subtraction is a neat trick. We can also use this idea to multiply numbers. I’ll just work this one out for you. I’m going to do the following problem:

( 52466789 )( 6542231) I actually have the option of doing this with logs instead of just straight away. Remember, if I’m using a table of logs, I’m probably (well, definitely, really) losing accuracy, but I’m gaining time, because it’s much easier. Let’s do it. First I’ll find the log of both numbers, and get:

log ( 52466789 ) ≈ 7.7198844859204 and log ( 6542231) ≈ 6.8157258745881 So now the question is just how accurate I want to be. If I go with 8 decimal places for my log approximations, I’d have:

( 52466789 )( 6542231) ≈ (10 7.71988449 )(10 6.81572587 ) = 1014.53561036 All I have to do now is add the exponents and then look that number up in my log table. My calculator gives me the result in scientific notation, which is sort of annoying. But when I asked it to subtract the approximation from the exact value I get 401540. That seems like a big difference. But calculating the percent error, approximation − actual (100 ) , actual my calculator says the error is 0.00000011698% , which is pretty much nothing. This number is so big that half a million either way doesn’t really make a difference.

Cool Log Things

10

www.turksmathstuff.com