How to figure out Logs?

[USArmyDoc]

I hate math and I am not to good in it. How the heck do you calculate logs?

For example: Log 3

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[mandy_848]

i don't know when they'd ever actually expect you to calculate the log of 3

it's mainly log10s that you need to worry about for pH and [H] and in those cases, you pretty much just look at the exponent

ex: if [H] is 10 to the exponent -2, the pH is 2

ex: if the pH is 3, the [H] is 10 to the exponent -3

i don't know if that makes sense, but i don't know how to explain it any other way. hopefully someone else can clear this up in a less confusing way

 

[USArmyDoc]

I understand that. However, I am doing my EK book at it asks you for the log of 3.

 

[mandy_848]

really?? i've never came across an mcat question where it wasn't based on log10s. i don't know how they'd expect you to know the log of 3.

 

[DrBowtie]

Edit: I was confused earlier. Below is correct.

log (3) = .477
10^.477 = 3

since .477 ~ .5
and 9^.5 = 3

you can approximate the answer.

So in recap,
log(x)=y
10^y=x

 

[USArmyDoc]

Brett,
You are the man!

 

[TIGIBedHead]

10^3 = log 3
to find any log just do the inverese log which is 10^n

I think....it sounded good while i was writing it. I will confirm later.

I haven't done math in a while, and was also having trouble with this. If y= log x, then x = 10^y. So if y= log 3, 3 = 10^y. Y has to be between 0 and 1. I don't think they would expect you to find the actual value.

Also, if you know that log 1 = 0, and log 10 = 1, then log 3 has to be in between those two values.

 

[scrubsRcomfy]

Also, if you know that log 1 = 0, and log 10 = 1, then log 3 has to be in between those two values.

yeah, that's all you really need. if you get a pH of 3.6, you know that the [H+] will be Y x 10^-4, where Y is some number between 1 and 10. once you get that far, the answer's usually apparent.

 

[DrBowtie]

Sorry about my earlier post. I was confused. I edited it to contain the correct info.

 

[rs76]

OK here is the best way to approximate the logs.
You wont find this in many places but here is the secret to approximate logs.

Just remember 248 and 369

log 2=0.2 log 4=0.6 log 8=0.9

and log 10=1

You can now approximate any logs encountered on the MCAT.

Good Luck

 

[kimijohni]

OK here is the best way to approximate the logs.
You wont find this in many places but here is the secret to approximate logs.

Just remember 248 and 369

log 2=0.2 log 4=0.6 log 8=0.9

and log 10=1

You can now approximate any logs encountered on the MCAT.

Good Luck

LOg 2= .3 not .2

 

[rs76]

Thanks for the correction. Typed it in wrong

LOg 2= .3 not .2

 

[Mister Pie]

Nonethless, it's an EXCELLENT way of approximating logs. Thanks for sharing it!