math question

[deewhit12]

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how do you take the log of a number without a calculator. for example, log of 0.889.

 

[CATallergy]

how do you take the log of a number without a calculator. for example, log of 0.889.

I assume you want a rough trick

First, remember that log (A x B) = log A + log B
(and of course, log 10 = 1 and log 1 = 0)

Now, some memorization - remember the range .16 to .105

(log 2) / 2 = .151
(log 3) / 3 = .159
(log 4) / 4 = .151
(log 5) / 5 = .140
(log 6) / 6 = .130
(log 7) / 7 = .121
(log 8) / 8 = .112
(log 9) / 9 = .106

Notice that the range starts around .16 and ends around .105, with midpoint around log 6.5.

Now, whenever you see a problem, factor out the 10^x part, and then use this approximation.

In your example, log 0.889:

1) log (0.889) = log (10^-1 x 8.89) = log 10^-1 + log 8.89 = -1 + log 8.89

2) now 8.89 is at the high end of our scale (from .16 - .105), so maybe we'll guess .107

3) -1 + log 8.89 = -1 + (8.89)*(.107) = -0.049

calculator shows log 0.889 = -0.051


Now, if you want to calculate a log, not in base 10, there is an easy conversion formula that is hard to type here, but I'll try:

log (base b) x = (log x) / (log b)

using b = e (since typing ln is easy!), an example:

ln 0.899. (e = 2.72)

1) ln 0.899 = log 0.899 / log 2.72

2) 2.72 is at the beginning of the range, so using our trick, assuming .155 for our trick value, log 2.72 = 2.72*.155 = 0.422 (calculator says 0.435)

3) thus, using our estimates, ln 0.899 = -0.049/0.422 = -0.116

calculator says: -0.118

how do you like this trick? I came up with it up in high school, but I'm sure someone else has as well.

 

[deewhit12]

🙂

 

[Rofeh20]

That was an impressive post there, but the AAMC will never ask you a math question that requires that level of calculation.

Here's what you need to know about logs:

1) Log (1) = 0
2) Log (10) = 1
3) Log of any number b/w 1 and 10 = fraction
4) Log of any number less than 1 = negative number

For example, let's say you need to find the pH of a solution whose concentration of H+ ions is 6.3 * 10^-8.

The pH must be between 7 and 8 so you'll be able to narrow down your answer choices quickly just by picking that up. The next step is to focus in on the "6.3" term. Log (6.3) is going to end up being some number between 0 and 1, and since 6.3 is closer to 10 I would estimate that log (6.3) is probably around 0.7 so my prediction would be 7.3.

It turns out that log (6.3) is equal to 0.799 so the actual pH will be 7.2. The AAMC will not list answer choices that are very close to 7.2 though. For example, the answer choices for the question I posed above would probably be:

A) 6.7 ([H+]=5.1*10^-7)
B) 7.2 ([H+]=6.3*10^-8)
C) 7.8 ([H+]=1.6*10^-8)
D) 8.1 ([H+]=1.2*10^-9)

These four answer choices may seem like they're relatively close, but if you take a look at the corresponding H+ concentrations for each answer choice you can see that they're not. As I noted above, once you know that the H+ concentration is to the -8 power you can rule out answer choices A and D; and then it's just a matter of looking at the mantissa and approximating. Since 6.3 is closer to 10 than 1 you know that the pH must be closer to 7 than 8.

PS: I am presuming that everyone here knows log rules...specifically, that is log (xy) = log(x) + log👍, thus for the question above the following simplifications occur:
-log(6.3*10^-8) = -log(6.3) + -log(10^-8).
The -log(10^-8) is equal to 8, thus we end up with 8 - log(6.3).
Notice that once you get the concentration in scientific notation, the pH is simply going to be exponent and one less than the exponent. Thus, since I saw it was 10^-8 I knew that the pH was going to be between 7 and 8.


That is really all the approximating that is necessary. Algebraic manipulations can also be tested, and you do need to know that natural log has a base of e instead of 10, but the MCAT will never ask you to be precisely calculate the log of any non-standard number.

 

[rls303]

That was an impressive post there, but the AAMC will never ask you a math question that requires that level of calculation.

Here's what you need to know about logs:

1) Log (1) = 0
2) Log (10) = 1
3) Log of any number b/w 1 and 10 = fraction
4) Log of any number less than 1 = negative number

For example, let's say you need to find the pH of a solution whose concentration of H+ ions is 6.3 * 10^-8.

The pH must be between 7 and 8 so you'll be able to narrow down your answer choices quickly just by picking that up. The next step is to focus in on the "6.3" term. Log (6.3) is going to end up being some number between 0 and 1, and since 6.3 is closer to 10 I would estimate that log (6.3) is probably around 0.7 so my prediction would be 7.3.

It turns out that log (6.3) is equal to 0.799 so the actual pH will be 7.2. The AAMC will not list answer choices that are very close to 7.2 though. For example, the answer choices for the question I posed above would probably be:

A) 6.7 ([H+]=5.1*10^-7)
B) 7.2 ([H+]=6.3*10^-8)
C) 7.8 ([H+]=1.6*10^-8)
D) 8.1 ([H+]=1.2*10^-9)

These four answer choices may seem like they're relatively close, but if you take a look at the corresponding H+ concentrations for each answer choice you can see that they're not. As I noted above, once you know that the H+ concentration is to the -8 power you can rule out answer choices A and D; and then it's just a matter of looking at the mantissa and approximating. Since 6.3 is closer to 10 than 1 you know that the pH must be closer to 7 than 8.

PS: I am presuming that everyone here knows log rules...specifically, that is log (xy) = log(x) + log👍, thus for the question above the following simplifications occur:
-log(6.3*10^-8) = -log(6.3) + -log(10^-8).
The -log(10^-8) is equal to 8, thus we end up with 8 - log(6.3).
Notice that once you get the concentration in scientific notation, the pH is simply going to be exponent and one less than the exponent. Thus, since I saw it was 10^-8 I knew that the pH was going to be between 7 and 8.


That is really all the approximating that is necessary. Algebraic manipulations can also be tested, and you do need to know that natural log has a base of e instead of 10, but the MCAT will never ask you to be precisely calculate the log of any non-standard number.

Hi,
Thanks for that post, I've gotten most of it down already.

For any of the recent takers or those who have taken recent prep courses, is this all the knowledge that will suffice?

I'm HORRIBLE at mental math, so I hope I will have it to a minimum on my exam.

 

[Isoprop]

I assume you want a rough trick

First, remember that log (A x B) = log A + log B
(and of course, log 10 = 1 and log 1 = 0)

Now, some memorization - remember the range .16 to .105

(log 2) / 2 = .151
(log 3) / 3 = .159
(log 4) / 4 = .151
(log 5) / 5 = .140
(log 6) / 6 = .130
(log 7) / 7 = .121
(log 8) / 8 = .112
(log 9) / 9 = .106

Notice that the range starts around .16 and ends around .105, with midpoint around log 6.5.

Now, whenever you see a problem, factor out the 10^x part, and then use this approximation.

In your example, log 0.889:

1) log (0.889) = log (10^-1 x 8.89) = log 10^-1 + log 8.89 = -1 + log 8.89

2) now 8.89 is at the high end of our scale (from .16 - .105), so maybe we'll guess .107

3) -1 + log 8.89 = -1 + (8.89)*(.107) = -0.049

calculator shows log 0.889 = -0.051


Now, if you want to calculate a log, not in base 10, there is an easy conversion formula that is hard to type here, but I'll try:

log (base b) x = (log x) / (log b)

using b = e (since typing ln is easy!), an example:

ln 0.899. (e = 2.72)

1) ln 0.899 = log 0.899 / log 2.72

2) 2.72 is at the beginning of the range, so using our trick, assuming .155 for our trick value, log 2.72 = 2.72*.155 = 0.422 (calculator says 0.435)

3) thus, using our estimates, ln 0.899 = -0.049/0.422 = -0.116

calculator says: -0.118

how do you like this trick? I came up with it up in high school, but I'm sure someone else has as well.

Way too much memorization IMO.

here's a much easier trick that works well enough for the MCAT (remember, you dont need to calculate EXACTLY the answer since this is a multiple choice test).

log 2 = 0.3
log 4 = 0.6
log 8 = 0.9

notice a pattern? logs of powers of 2 = the multiples of 3 * 0.1.

That, and memorize log properties: log A*B = log A + log B and you can do all the log problems on the MCAT.

ex: what is the pH of a 5*10^-3 M HCl soln?
answer:

pH = -log(5*10^-3)
= -(log(5*10^-3)
= -( log(5) + log(10^-3) )
= -(log(5) - 3)

log 5 is a little above log 4. So estimate it at 0.7.

= -(0.7-3)
= 2.3

 

[Isoprop]

or using OP's example:

for example, log of 0.889

log0.889 = log(8.99*10^-1) =
=log(8.99) + log(10^-1)
=0.95 - 1
= -.05

for the mcat, log8.99 = log 9 = 0.95 (a little bit above log 8 = 0.9)

calculator answer is -0.0462403083

 

[qwerty75]

I took the MCAT last year & you only have to know log base 10 - nuttin' else. I found good sites for review were http://www.physics.uoguelph.ca/tutorials/LOG/ and http://mcatprep.yolasite.com/qq11. BTW, i got a 14 on PS.

 

[BrownRoach]

Also check wikipedia for concise info on any topic.

 

[NYR56]

Way too much memorization IMO.

here's a much easier trick that works well enough for the MCAT (remember, you dont need to calculate EXACTLY the answer since this is a multiple choice test).

log 2 = 0.3
log 4 = 0.6
log 8 = 0.9

notice a pattern? logs of powers of 2 = the multiples of 3 * 0.1.

That, and memorize log properties: log A*B = log A + log B and you can do all the log problems on the MCAT.

ex: what is the pH of a 5*10^-3 M HCl soln?
answer:

pH = -log(5*10^-3)
= -(log(5*10^-3)
= -( log(5) + log(10^-3) )
= -(log(5) - 3)

log 5 is a little above log 4. So estimate it at 0.7.

= -(0.7-3)
= 2.3

This is exactly what you need to know. Learn this post and nothing else, unless you're entering a math competition. For the record, the choices on the AAMC/MCAT are generally pretty far apart. EKs were moderately far apart, but less than the real thing and Kaplan had some ridiculous calculations in their practice questions.