MCAT Math Vid 8 - Logarithms and Negative Logs in pH and pKa Without A Calculator
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- เผยแพร่เมื่อ 7 มิ.ย. 2014
- Leah4sci.com/MCATMath presents: MCAT Math Without A Calculator Video 8 - Solving MCAT logarithm calculations without a calculator
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This video is part 8 in my MCAT Math Without a Calculator video series. This video teaches you how to solve the potentially impossible MCAT logarithm questions without a calculator by showing you what to look for and numbers to recognize for simple and fast conversions.
This video also demonstrates the log trick by showing you how to solve an OH- concentration to pH question.
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▸MCAT Chemistry Series + Cheat Sheets Leah4sci.com/MCATchemistry
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Just in case anyone wants to know a faster trick: Kaplan teaches the Ninja Trick. Anytime you are given a number and have to find the 'p' of it (AKA log it).
Trick: (exponent - 1).(10 - mantissa)
*That dot is a decimal place; mantissa = number in front of the x 10 ^#
(4-1).(10-2.3)
4-1 = 3
10 - 2.3 = basically 8
pOH = 3.8
To find pH: 14 - 3.8 = 10.2
*This answer is more accurate than Leah's as well
This has helped me greatly, because math is not my strong suit, and I am terrible at rounding or thinking about how numbers fit into things. This is also crazy fast and will save you a lot of time. I hope this helps someone who is on the log-struggle-bus.
However, Leah's video is great for understanding the concept of Logs as a whole!! Props.
sometimes ur method is not accurate take log 3*10^-2
ur method shows 1.7 but her shows 1.5 which is correct....
just a suggestion
that's true, but for the MCAT, still close enough. I just shared Kaplan's method for people like me who reallyyyyyy struggled, because I still don't fully understand her teachings, which isn't helpful for me.
Thanks for the suggestion. It's always good to have alternative methods for different learning styles
This way is soo much more easier to understand and remember!
dude you saved my life!!!!! thanks so much!!! logs are so much easier now
This video literally saves lives
It does fr sure
Aww thanks!
-log(m x 10-n) ≈ n - 0.m this is also a good method
Bless your soul
You literally saved me thankssss a tonnnes❤❤❤❤✨😭
That doesn’t work
I haven't heard of this one before
I know this is a less accurate method, but it's quicker.
If you see -log(1.5x10^-4) ... I just do 4.0 - .15 = 3.85, which is pretty close to the real answer of 3.82.
I know this becomes less accurate with certain numbers, but you don't have to be THAT close on the MCAT.
thats how I do it too, I wish she would have mentioned it so more people would know about it.
Yeah I read that in a book as well and was going to comment that. I'll stick to that method since its much easier. Hopefully it will be enough for the mcat.
9.3X10^-7..... way off. I wonder if there is a way to tell when it will be inconsistent
@@josephhicks9454 Wait... how is the alternative methodology "way off"? 7-0.93 = 6.07. Calculator says it's 6.03. That's very close and so much faster.
@@josephhicks9454 6.07 (estimate) vs 6.03 (actual) is close enough for me
In approximately 7 minutes, you helped me to overcome 1 problem that I have been grappling with for 3 days, I tried so many videos including Khan Academy (which is also great BTW) and scoured the internet for help till I found your video, thanks and I will be watching your other videos, taking the MCAT Jan 2015.
battlerook good luck on your MCAT!
Tnx )
battlerook I completely agree. Simple and to the point video that was extremely helpful! Wish I would have found this much sooner! Thanks Leah4sciMCAT
How did your mcat go?
How did the MCAT go?
This was definitely a helper especially for pH indicators on the MCAT! Thank you :)
Yay, glad to hear it!
Wow i needed this video bad lol. I relied heavily on my ti-83 through out college and now its cold turkey into the mcat with no calculator. Thanks for the math trick Leah =) your very good at explaining things
+Raymond Dion you, me and everyone else ;)
Glad you found the video helpful
Easier way to do negative logarithms: if you have... -log (y x 10^z ) = (|z|-1).(10-y) you get your number to an approximate decimal! So if you were to try the same number: -log (2.3x10-4) = (4-1).(10-2) = 3.8! (Just remember not to use the negative value in the exponent)
Thanks for sharing! Use whatever is easier/quicker for you.
it is nearly correct nice trick
Wow this is a great trick, thanks!!
Omg I was super depressed because I felt stuckthis helped me so much! Thank you
glad to help!
You are a saint for this, thank you! This will help immensely on the DAT!
Glad it was helpful!
This was super helpful. Thank you Leah
You're very welcome!
Thank you this video is very helpful. I am currently studying for the MCAT.
glad to help
Thank you for your videos, they are awesome!
Glad you like them!
Amazing, useful for the DAT. Thank you!
You're welcome!
The logic used is really great. I appreciate this.
Nice
You're welcome!
This is extremely helpful. Thank you 😊
You're very welcome!
U saved my life :)) thank u so muchhh
You're very welcome! Glad I could help!
Omg Leah! Thank you 😊
You're so welcome!
amazing.....thank you so much!
lirim sopaj you're welcome
I don't even know what MCAT EXAM is'but i have been struggling hard wid my prep for the boards.This video could solve my problem in one go.Thank u very much!!!!!
Wow, you're so welcome!
This is incredible!
Glad you like it!
This is everything thank you so much
You're so welcome!
Really great video!
hey, thanks!
🔥 Thank you.
you're welcome!
the vedio is really helpful and useful.thank u😊☺
+Venkatesh K Raja thank you
fantastic explanation...saved me hours on google...
Awesome! Glad you liked it!
Wow, incredible ❤️❤️❤️❤️
Thanks!
Thank you!
you are welcome
this was peeerfect thank you!
Yvette Rodriguez thank you
Thanks a lot!!
You're welcome!
This video is very informative. If I can make one suggestion, when you are giving the explanation and then the answer like when you do at 8:55... if you want to help students following along, I would write out these steps. For me, I had to watch this video twice and write out what you were explaining in order to understand how you got the answer.
this!!! i was so lost and confused when she was explaining it
Thank you so much for the feedback! That's a great idea and I will try writing out steps on the screen
THANK YOU!!!!!!
You're very welcome!
yo this is fooking awesome
Thanks
For me and probably everyone else it is just easier and more efficient to just remember and commit to memory log 1 through log 9. Thanks for the video.
BAdBrAd Memorization can lead to panic and confusion. Memorization based on logic is the better way to go
Sound logic. Agreed. :)
Thisssss issss POWER!!!!!!
'Cause knowledge is power!
great video should do videos for the DAT too..
thanks! one step at a time though. Much of the DAT sciences do overlap but the math is much more intense
thank you thank you thank you soo much
You're welcome, you're welcome, you're welcome so much! :)
Can you explain the 4.5 x10^-3 problem that you did. How did you go from 4.5 -> 1x10^-3 and where is the 10x10^-3 portion coming from?
See previous comments that answer this very same question. I realize that the video isn't very clear and plan to record a revised version soon.
I love you 😘 you are a life saver
+Girlie Stylistic Thank you! Glad to help
I watched the video multiple times to try and understand but I was/am very confused at the end and how you went from 2.3 to 1 to 4 and all that in between the answer.
Nicole Merkelbach same I am very lost
At which specific point in the video?
THANK YOU
You're welcome!
Confused how you rounded down from 4.5 to 1. I read in other comments that you had previously answered this question and were planning on making a revised video, but I haven't seen either. Can you please explain?
I haven't posted that video just yet, sorry for the confusion
Hey Leah,
I'm a bit confused on your method, I found another hack, but I'm trying to find a way for it to work alongside yours!
Any tips/hacks or tricks for solving antilogs with a number OTHER than 1?
For ex: -log(2.5X10-4) or -log(7.1X10-4)
I like the trick you used which blew my mind; counting the 'numbers' after the 1 to get an approximation; for ex: log(127) = approx 2, because if you count both 2 and 7, this is 2 numbers after the 1, therefore approx 2.
HOWEVER, if the log isn't beginning with a 1 (ex: log(127)), and instead a 4 or 7, (for ex: log (403) or log(7992) how do we ensure it is still accurate using this method?
Using your method Jesse, and continuing using the examples log(403). Would you then round up to log(1000), then count 3 numbers after the 1(that being three zeros), therefore we estimate the answer to be approx 3. Or roughly guestimate; log(403) there are two numbers after the 4, therefore (using both my trick and your trick) estimating it is between 2-3 applying both rules.
Or for log(7992), similar method for solving; round up to log (10000), then there is 4 numbers after the log, so approx 4? Or again, applying both of our rules to solve, log(7992) there are 3 numbers after the 7 therefore the answer should be somewhere approximately between 3-4?
Does my question make sense?
Thank you all in advance for your time and efforts!
There are many ways to approach this concept. The key is to find what is
1- fast enough
2- close enough
3- works BEST for you
If you like the method you listed and it works out, go for it. I personally like tricks that are quick and simple, otherwise it's too much of a distraction. Did you see my simplified approach here? th-cam.com/video/Pn4cp-ai_C8/w-d-xo.htmlsi=FeM6Yj2it9MgBPRJ
how do you determine if the answer is negative or not? Ex. log(0.054)
If you take the log of a number greater than 1 the answer is positive. If you take the log of a number less than 1 then it's negative
This is great. Now let's say we wanted to calc. the unknown concentration of [H30+] using the pH in this case I know you would just do 10^-pH. I understand this part, however, how would you calc. for example [H30+] = 10^-2.34 to find the concentration WITHOUT using a calculator.
Sure. We might not be able to find the EXACT number without a calculator, but we can make a pretty good estimation by assuming the answer will fall within the range of 10^-3 and 10^-2. We can narrow that range even further by memorizing a set of decimal indicators as I share in my antilog video at leah4sci.com/solving-antilog-mcat-questions-without-a-calculator/. Remember on the MCAT: close enough is good enough. We only need to choose the right answer from a set of multiple choice options.
0=1, 1 = 1.25, 2 = 1.5, 3 = 2, 4 = 2.5, 5 = 3, 6 = 4, 7 = 5, 8 = 6, 9 = 8, 10 = 10. This is a repeating cheat sheet i use. Its not perfect but it gets the job done.
Thanks for sharing!
What would happen if you were taking a positive log such as in the Henderson-Hasselbach equation? If you were taking a log of (.4/.6) How would you do that mentally? I know you would have to split it into log(.4) - log(.6) but I am not sure how you would do that.
this sounds far beyond the scope of a typical MCAT question
great
Thanks!
Hi Leah, not sure if you still see this, but how would i apply this rule to positive logs? I tried doing the mental calculation of how 3 would be around 0.5... how 5 would be around 0.3 if these were negative, etc. It doesn't work the same for positives, is there something I missed in the video? A response would be GREATLY appreciated!
Thanks for your question! No, you didn't miss anything in the video. Your answer would require a completely different video, and while it's in the plans, I don't expect to get around to it any time soon.
Ur the bomb I love u 💣
Aww, thanks!
omg do you know how much I was stressing cuz of this! thank you!!
+Kerwyn204 so glad to help you.
Still didn't get it. Lost in translation from round down 4.5*10^-3 to the 1*10^-3 and 10*10^-3 aka 1*10^-2, then some gibberish about .5, .3,.1 I have no concept anchor at that point of the lesson. 😕😭😞
the exponent is the whole number for a log. So 1/1000 = 3 negative zeroes, which is -3.0, while the pocket change left over is on a sliding scale.
Here's an extended table:
0.01 0.0125 0.015 0.020 0.025 0.030 0.040 0.050 0.060 0.080 0.100 log -2 to log -1
0.10 0.1250 0.150 0.200 0.250 0.300 0.400 0.500 0.600 0.800 1.000 log -1 to log 0
1.00 1.2500 1.500 2.000 2.500 3.000 4.000 5.000 6.000 8.000 10.00 log 0 to log +1
10.0 12.500 15.00 20.00 25.00 30.00 40.00 50.00 60.00 80.00 100.0 log +1 to log +2
100.0 125.0 150.0 200.0 250.0 300.0 400.0 500.0 600.0 800.0 1000 log +2 to log +3
1000 1250 1500 2000 2500 3000 4000 5000 6000 8000 10000 log +3 to log +4
and each step is about 0.1 log, thus the "3" position is 0.5 log units, the "8" position is 0.9 log units.
shintashi Thanks for trying but that long chart didn't help me understand what she was trying to show in the lesson. I understood u up until your sliding chart. 😔
I'm sorry, but I don't offer tutoring through TH-cam comments. For help with this and more, I recommend joining the MCAT study hall: join.mcatstudyhall.com/
@@Leah4sciMCAT Hi Leah, thank you for the video, another quicker way of solving for the pH is taking taking the exponent and subtracting, example: 2.3x10^-4 you could do 4-.23= 3.7 which is pretty close for the MCAT.
For the -log(4.5 x 10^-3) question, (you didn't get the chance to solve it in the video) was the answer 2.5? I just want to confirm. TYSM!
No, although I don't give a final answer for that calculation, I do say that it's NOT going to be halfway between my pH's of 2 and 3. In order to determine a more exact number without using a calculator, we have to memorize the values shared around the 8:00 mark in the video. 4.5 x 10^-3 is close to 5 x 10^-3, and so we know the decimal point is going to be closer to 0.3 in our answer. That makes our final answer somewhere close to 2.3 rather than 2.5.
@@Leah4sciMCAT Thank you!
here for gamsat - thank you
You're welcome!
do you have a video on natural logs?
Not yet
How would a reverse problem work? Let’s say h+ concentration is given and u need find the pH
I'm sorry, but I don't provide tutoring through TH-cam comments. For help with this and more, I recommend joining the MCAT study hall. Full details: join.mcatstudyhall.com/
@5.50,
I dont understand what is meant by the 'round down' and 'round up'. Can someone please explain?
I realize that this trick is not as intuitive to follow, that's why I created an updated simpler trick here: th-cam.com/video/Pn4cp-ai_C8/w-d-xo.htmlsi=FeM6Yj2it9MgBPRJ
almost cried bc i cant do math in my head you made a clueless person understand in 5 minutes
Awww, I'm so happy to hear that I help you understand!
damn girl, well done
Thank you for your kind words
What if the problem asks -log(0.20)? I tried the shortcut but I get between 1 and 1? Is there something I am missing? On the calculator the answer says .698
I'm sorry, but I don't offer tutoring through TH-cam comments. For help with this and more, I recommend joining the MCAT study hall: join.mcatstudyhall.com/
-log 0.2 is -log 2/10 = log 10-log2 = 1-0.301 = 0.699
Thank you!! :)
Can someone explain why at 9:02 she rounds up to 10 resulting in her getting 3 in her range?
Sure! 2.3 x 10^-4 is somewhere between 1 x 10^-4 and 10 x 10^-4. We already know that the -log of 1 x 10^-4 is 4. For the other side of the range, we know that 10 x 10^-4 is EQUAL TO 1 x 10^-3. And the -log of 1 x 10^-3 is 3. That means our pH is going to be somewhere between 3 and 4.
CAn someone pls show me how to do 10^-1.16? (trying to find [H] from pH but can't find any tutorials).
On the MCAT 'close enough is good enough' you would choose and answer that is close to the value of 10^-1 knowing that if given one higher and one lower you go with the value ever so slightly approaching 10^-2 (something slightly smaller than 0.1)
How did you get the -2 at 6:31?
I'm sorry, but I don't offer tutoring through TH-cam comments. For help with questions like this and more, I recommend joining the MCAT Study Hall. For more details visit join.mcatstudyhall.com/ or contact me through my website leah4sci.com/contact/
badass
Thanks!
why did you take the smaller number (3)?
actually io think i understand... do we take the smaller number because taking the bigger number would put the decimal out of range?
If you're speaking of the last example, we want our pOH value to be somewhere between 3 and 4. The decimal number needs to fall somewhere in between those two whole numbers. Saying 4.5 would put it out of range, yes.
What about - log base 10?
What is your question exactly? This video does address negative logs with a base of 10. Could you be more specific?
Why didn't you list what the -log values of 2, 4, 6, 7 and 9 were?
I have only listed those values that I believe are important for you to know in order to estimate logarithms quickly without a calculator on the MCAT. To reference my 'need to know' values, download my MCAT Math Cheat Sheet at leah4sci.com/mcat-math-study-guide-cheat-sheet/
I LOVE U
thanks!
Starts @ 2:00 (If in a hurry lol)
Thanks for watching and commenting!
I think it's important to not that this method only works with numbers less than 1.
Yes, this method is geared towards solving negative logs in pH and pKa problems on the MCAT. The negative log of a number greater than 1 would yield a negative pH value, which goes against our knowledge of the pH scale ranging from 0 to 14. In other words, taking the negative log of a number greater than 1 doesn’t apply here. Thanks for the comment!
Wot? This is very complicated. I tool grade 11 university chemistry and I did not do this.
What was complicated? The math or the explanation?
😢😰😰😰😰😭😭😭😭😭😭l am very sad because I couldn't understand _log questions..
I'm sorry that you had trouble understanding them. For more help with this, I recommend joining the MCAT study hall. Full details: join.mcatstudyhall.com/
This really isn’t good enough
I'm sorry you feel that way. What's missing?
What's an mcat?
It's the medical school admissions test given primarily in the US and Canada
You're really supposed to be able to estimate decimal/fractions of a -log? LOL
Yes, especially as it applies to pH and pKa problems. Calculators are NOT allowed in taking the MCAT, so test takers must have a fairly good grasp on various mental math skills. For more on MCAT Math, watch my entire series at leah4sci.com/mcat/mcat-math-without-a-calculator/.