Calculating Logs on the DAT

[BlackoutRuse]

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I don't know how to calculate logs and am getting confused.

I have to calculate the -log(3.2x10^-3)

Can someone give me a clear rundown on how to do that

My practice thing says that an estimation between 2 and 3 is good enough to find the answer for this problem, as the answer for this question is 2.5.

 

[Chemboy23]

Ok, so on these, you just take the expontent (make it positive and subtract 1) and that gives you your whole number. Next, to get the decimal that follows, you have to sort of memorize a little table as follows....

Basically, the exponent becomes your whole number and the whole number (3.2 here) becomes the decimal.

1.0=1
2.0=0.69
3.0=0.52
4.0=0.39
5.0=0.30
6.0=0.22
7.0=0.15
8.0=0.09
9.0=0.04

So, let's say we have -log(5.7 x 10^-7)

We take our negative exponent and subtract one, giving us 6.

We know it's going to be 6.something...

To get the rest, look at your table and estimate a value...5.7 gives us around ~0.24...

So, your answer will be scarily close to 6.24!!!

Hope this helps a little...I had to memorize it too lol!!

Good Luck

 

[dentdavis]

this was the best explanation ever! thank you so much..this helped tons!

 

[Chemboy23]

YW...anytime! Good Luck!

 

[DQLEUCD]

Ok, so on these, you just take the expontent (make it positive and subtract 1) and that gives you your whole number. Next, to get the decimal that follows, you have to sort of memorize a little table as follows....

Basically, the exponent becomes your whole number and the whole number (3.2 here) becomes the decimal.

1.0=1
2.0=0.69
3.0=0.52
4.0=0.39
5.0=0.30
6.0=0.22
7.0=0.15
8.0=0.09
9.0=0.04

So, let's say we have -log(5.7 x 10^-7)

We take our negative exponent and subtract one, giving us 6.

We know it's going to be 6.something...

To get the rest, look at your table and estimate a value...5.7 gives us around ~0.24...

So, your answer will be scarily close to 6.24!!!

Hope this helps a little...I had to memorize it too lol!!

Good Luck

Your calculations is off. -log(5.7 x 10^-7) should be about - 6.4 or so. Easiest way to do it is whatever power it is to the base 10 is that number if it's 1 x that base 10 number. So in this example the answer if it's 1x 10^-7 is 7 and since the number is larger then 1, it should be lower then then 7 so about -6.4. also if your taking a (-) to that log, u add the - to it so it's about -6.4.

so it's more then 1 it should be a little less then that power if it's a lot bigger then 1 then it's a lot less.

 

[Chemboy23]

Sorry my friend, but you are wrong by about 0.15...when I put it in the calculator, I get 6.244125144...try figuring something out before you accuse someone of spreading misinformation...thanks! 😀

 

[Chemboy23]

---->BlackRose...yes your estimate will probably be sufficient on the real DAT. I mean, if you think about it...we're not allowed a calculator on this section so they can't get too technical.

Just for kicks...

3.2 x 10^-3=~2.49-2.50ish

Hope this helps you a little!👍

 

[Troyvdg]

what about natural logs?
for equations such as DG=(RT)(ln)(Keq) or Ecell= E*cell-(RT/nF)(lnQ) ?

 

[DQLEUCD]

Sorry my friend, but you are wrong by about 0.15...when I put it in the calculator, I get 6.244125144...try figuring something out before you accuse someone of spreading misinformation...thanks! 😀

No need for calculator or memorize all those numbers when you can do a trend. As it gets larger then 1, it's going to get less then whatever the exponent is. The larger it is the farther it is from the exponent number. Unless you want to memorize log of 1 - 9, my way of thinking is a lot easier to grasp.

 

[Jackson-Ave-Dental]

When you see log problems, it should be instant free points 😀
Just have to memorize this...
log 2 = .3
log 3 = .5
log 4 = .6
log 5 = .7
log 6 = .8
log 7 = .85
log 8 = .9
log 9 = .95



As an example: -log(3.2x10^-3) **approximate 3.2 as 3
= 3 - log 3.2
= 3 - .5
= 2.5

As another example: -log(5.7 x 10^-7) **approximate 5.7 as 6
= 7 - .8
= 6.2

Another example: -log(8.1 x 10^-5)
= 5 - .9
= 4.1

Another example: -log(6.9 x 10^-2)
= 2 - .85
= 1.15

Do you see the trend here?
All you have to do is take the exponent of 10 and subtract the log of the first number.
-log(3.2x10^-3)

3 - log 3.2