10^x without a calculator

[maxx52188]

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In order to work out the change in pH, we have to first calculate the initial concentrations
of acetic acid and sodium acetate. We already know from Table 1 that the concentration of
sodium acetate is 3M. Plugging this into the Henderson-Hasselbalch equation, we can work
out the concentration of acetic acid. The Henderson-Hasselbalch equation is as follows:
pH = pKa + log [A–]/[HA], where [A–] is the acid concentration of the conjugate base, and [HA] is
the concentration of the acid. Plugging our values into this equation, we get 5.2 = 4.74 + log
[3]/[HA]. The concentration of acetic acid, therefore, works out to be 1M. In a 100mL solution, the
number of moles of sodium acetate and acetic acid works out to be 0.3 and 0.1 moles, respectively


without a calculator, how in the hell did they know that 10^.46 is about 3?

 

[chiddler]

unless there's a better or faster way to do this that i'm unaware of, I suggest you compile a list of powers and such:

and study the pattern and memorize it.

 

[SaintJude]

No idea. I don't have TBR, but I'm already guessing you're using that. Along the way, I've seen that this forum suggests you memorize certain values

log 2= 0.3
log 3= 0.47
log 5= 0.6
log 8= 0.9

I can remember these values, just by noting that the answer of all of these small logs are usually the decimal of that value + 0.1 decimals extra." Don't believe that? Check log 7. So the log 2-8 should be hopefully easy to remember.

Edit: Looking at what I've typed, I realize I didn't explain the trend well, but I hope you can visually see the (+0.1) I'm talking about. Sorry.

 

[aSagacious]

On the real exam you will only have to do very rough calculations. For this problem there would probably be only one answer choice between 1 and 10. Knowing that 10^0 is 1 and 10^1 is 10 you'd be able to eliminate the rest. Don't waste your time memorizing trivia like this.

 

[MedPR]

In order to work out the change in pH, we have to first calculate the initial concentrations
of acetic acid and sodium acetate. We already know from Table 1 that the concentration of
sodium acetate is 3M. Plugging this into the Henderson-Hasselbalch equation, we can work
out the concentration of acetic acid. The Henderson-Hasselbalch equation is as follows:
pH = pKa + log [A–]/[HA], where [A–] is the acid concentration of the conjugate base, and [HA] is
the concentration of the acid. Plugging our values into this equation, we get 5.2 = 4.74 + log
[3]/[HA]. The concentration of acetic acid, therefore, works out to be 1M. In a 100mL solution, the
number of moles of sodium acetate and acetic acid works out to be 0.3 and 0.1 moles, respectively


without a calculator, how in the hell did they know that 10^.46 is about 3?

If this is TBR, they previously taught you that log2=0.3 and log3=0.48. Algebraically, "x=10^.46" is the same as "logX=0.46"


Also, you can memorize those patterns like chiddler suggested, or you can handle it how TBR suggests (I personally really like this method because it took my comprehension of logs and 10^x from 0 to MCAT-necessary-level).

TBR says to memorize that log2=0.3 and log3=0.477. Also, you should know that logA*B=logA+logB. Memorizing only those 3 things, you can solve any log problem on the MCAT.

For example, log8. 8=2*2*2, so log8=log2+log2+log2 =0.3+0.3+0.3 = 0.9

In chiddler's example.
-log4*10^-10 = (-log10^-10)-(log4) = 10 - (0.3+0.3)= 9.4
-log8*10^-10 = -log10^-10 + -log8 = 10-.9 = 9.1


Obviously prime numbers (like 7) are a little more tricky, but you know that log6=0.77 and log8=0.9, so log7 must be halfway between there, which is about 0.84.

On the real exam you will only have to do very rough calculations. For this problem there would probably be only one answer choice between 1 and 10. Knowing that 10^0 is 1 and 10^1 is 10 you'd be able to eliminate the rest. Don't waste your time memorizing trivia like this.

I've read in various sources that the MCAT tests pH up to +/- 0.5 accuracy. Meaning, you might have to be fairly exact.

 

[maxx52188]

First off, EVERYONE'S information was helpful, and let me know there are some things to memorize.

But my buddy told me how he'd do it, and I was embarrassed after cause it was so easy....


10^.46 is roughly 10^.5

10^.5 is really the square root of 10

square root of ten is barely over 3 (3.16 to be precise)

 

[chiddler]

yes but you want a method that works for more than just 10^0.5 like say 10^0.8

 

[MT Headed]

Not really. 10^0.8 is somewhere between 3 and 10. That's good enough on the genuine MCAT. It's certainly more than what most test takers would know.

 

[dmf2682]

Not really. 10^0.8 is somewhere between 3 and 10. That's good enough on the genuine MCAT. It's certainly more than what most test takers would know.

This. But if you were interested there's games you can play.

If you want 10^.8 you can say that's close to 10^.75 or sqrt (10^1.5)

Know that 10^1.5 = 10*sqrt 10 = 33. Which is close to 36. So now take your sqrt and its around 6.

If you really need something like this though, you probably did something wrong.

 

[MedPR]

This. But if you were interested there's games you can play.

If you want 10^.8 you can say that's close to 10^.75 or sqrt (10^1.5)

Know that 10^1.5 = 10*sqrt 10 = 33. Which is close to 36. So now take your sqrt and its around 6.

If you really need something like this though, you probably did something wrong.

Alternatively, you know that log6 is about 0.77. So 10^0.8 is a little more than 6.