logarithm

[Farcus]

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is there a way to estimate the antilog of something? I know you can easily estimate regular logs.

 

[spooge]

Just use the laws of logarithms to solve...or if this does not help...will you give an example of a question?

 

[sashagrl259]

does this help?

antilog (log x) = 10 ^ (log x) = x

or you can say, the antilog = 10 ^ (the answer of log x)

 

[Farcus]

what I mean was if the pH of blood is 7.4 then what is the hydronium ion?
i need to do 10^-7.4

 

[omnione]

what I mean was if the pH of blood is 7.4 then what is the hydronium ion?
i need to do 10^-7.4

It's between 10^-7 and 10^-8, probably closer to 10^-7 given the exponential nature of the question. I highly doubt the PCAT would clump similar answers to the real answer so closely, so I think you should be able to estimate to the correct digit at least.

 

[Farcus]

but how would you estimate this though? I know 10^-7 is 1.0 * 10^-7 and 10^-8 = 1.0 * 10^-8, so if i were to guess I'd say 4.5 * 10^-8?

 

[Farcus]

heck an even better question, how do you find the log of 10^.65? I mean whole number easy yes but decimal numbers?

also how do you do the sqrt of a scientific number? I mean I can't seem to find any order that it goes by...

 

[Transformer]

heck an even better question, how do you find the log of 10^.65? I mean whole number easy yes but decimal numbers?

also how do you do the sqrt of a scientific number? I mean I can't seem to find any order that it goes by...

log of 10^.65 will not be on PCAT



Just in case it shows up, this is how you
solve log (10^.6500)

(10^.6500) = 4.466835922
log(10^.6500) = log(4.466835922) = 0.65 {calculator}

"For numbers with just 1 digit before the decimal point (1 to 9), the index will be 0."

Index = 0

The Mantissa is only found on the Log Table

"Use the numbers at the far left of the table to give the first two significant figures of the number. If there are more than two digits in the number, follow across the table - the column headings give the third digit of the number."

Look for 44 on far left of table and 6 on column side; this shows the Mantissa for this problem is between 6493 and 6503 (see log table)

4.4668 shows the true value is closer to 6503 than 6493 thus, one would estimate log(4.4668) = 0.6499 or 0.6500

Index = 0 and mantissa = 6499 or 6500

hence:


log(10^.6500) = log(4.466835922) = 0.6500


See following websites for better understanding:

http://www.scenta.co.uk/tcaep/maths/number/logtab/index.htm

http://209.85.141.104/search?q=cache:xKaHf35SSckJ:www.rain.org/~rcurtis/logs.html+Finding+the+mantissa+of+a+logarithm&hl=en&ct=clnk&cd=2&gl=us&client=firefox-a



Here is a website for antilog:
http://members.aol.com/profchm/log.html

 

[ffpickle]

log of 10^.65 will not be on PCAT



Just in case it shows up, this is how you
solve log (10^.6500)

(10^.6500) = 4.466835922
log(10^.6500) = log(4.466835922) = 0.65 {calculator}

"For numbers with just 1 digit before the decimal point (1 to 9), the index will be 0."

Index = 0

The Mantissa is only found on the Log Table

"Use the numbers at the far left of the table to give the first two significant figures of the number. If there are more than two digits in the number, follow across the table - the column headings give the third digit of the number."

Look for 44 on far left of table and 6 on column side; this shows the Mantissa for this problem is between 6493 and 6503 (see log table)

4.4668 shows the true value is closer to 6503 than 6493 thus, one would estimate log(4.4668) = 0.6499 or 0.6500

Index = 0 and mantissa = 6499 or 6500

hence:


log(10^.6500) = log(4.466835922) = 0.6500


See following websites for better understanding:

http://www.scenta.co.uk/tcaep/maths/number/logtab/index.htm

http://209.85.141.104/search?q=cache:xKaHf35SSckJ:www.rain.org/~rcurtis/logs.html+Finding+the+mantissa+of+a+logarithm&hl=en&ct=clnk&cd=2&gl=us&client=firefox-a



Here is a website for antilog:
http://members.aol.com/profchm/log.html

there are no calculators on the pcat, and it seems this calculation requires a calculator. I wouldn't learn all this mantissa index stuff since most of us don't have time for it.

i agree with omnione's approach to solving 10^-7.4 (you know it is between 0.0000001 and 0.00000001, closer to the former). can't think of an easier way.

to solve 10^-0.65, i would do some major approximations:

10^-0.65 = 1/(10^0.65)
Let's say 0.65 ~ 0.5
Then,
1/(10^0.5) = 1/sqrt(10)

sqrt(9)=3 so sqrt(10) is probably ~ 3.something

Since 1/3=0.333, Then 1/3.something = slightly less than 0.333, maybe 0.3?

But remember in the beginning we said 0.65 ~ 0.5, so we would get an even bigger denominator in 1/3.something (go back to see why). So reduce the 0.3 even more, to say about 0.24-0.29? Who knows (or gives a **$#&), but at least we have a range....

The actual answer is 0.224

 

[omnione]

Here's a document from my Pharmacy Calculations class that was a great help in estimating logs and antilogs for the purpose of acid/base ionization problems.