Math log help!
Started by fifi1314
Hey Fifi, did you & that other poodle mimi go for your afternoon walk & had treats afterwards 🙂?
For your problem, in general, if you have a log expression of the form:
m x 10^-n,
the ans will always lie between n and n-1, and if m is less than 2, the value is closer to n greater than or equal to three, the closer the value is to n-1 so let's consider each choice in turn:
a) -log(6.5 x 10^-5) = closer to n-1 = 5-1 = 4, since 6 is well past three
b) - log (2 x 10^-4) = since 2 is less than 3, its the n value or 4
c) log (2 x 10-4) = Same as above, except here we're taking the positive log, so you just get -4
d) log (2 x 10^4) = simply 4
Hope this helps! Let me know if you need help with anything else, & good luck studying!
- Donjuan
For your problem, in general, if you have a log expression of the form:
m x 10^-n,
the ans will always lie between n and n-1, and if m is less than 2, the value is closer to n greater than or equal to three, the closer the value is to n-1 so let's consider each choice in turn:
a) -log(6.5 x 10^-5) = closer to n-1 = 5-1 = 4, since 6 is well past three
b) - log (2 x 10^-4) = since 2 is less than 3, its the n value or 4
c) log (2 x 10-4) = Same as above, except here we're taking the positive log, so you just get -4
d) log (2 x 10^4) = simply 4
Hope this helps! Let me know if you need help with anything else, & good luck studying!
- Donjuan
Hey,
I tried to work thru those questions by ur method, got questions again...
Those '-' I threw into examples can be confusing.
1. can we just say all case are following log(m x 10^n) = n-1 when m > 2?
2. what's the answer of log(2 x 10^2)=?
3. Is -log (6x10^-2) = log(6x10^2)??? if not, how to deal with those '-'?
Thank you alot!!
I've just recently been reading more into applying to dental schools and taking the DAT... My first instinct when I read your question was to just type it into the old Ti 83, I could literally do this with my eyes closed after nonstop pH calculations for gchem2. But I just read that you can't use calculators on the DAT and I think I peed a little 
, no a lot. I'm so accustomed to just using the inverse log buttons and square roots and all that. I even use the calculator for addition if it involves 3 digits. I suppose I have a little less than a year to get my math machine in shape... crap.
, no a lot. I'm so accustomed to just using the inverse log buttons and square roots and all that. I even use the calculator for addition if it involves 3 digits. I suppose I have a little less than a year to get my math machine in shape... crap.For log(m x 10^n) you get (by one of the properties of logs --> google that if you don't know them) log(m) + log(10^n). By another property of logs that equals log(m) + n*log(10) and log(10) = 1 so you have log(m) + n.
First let me point out that m can never be negative since you can't take the log of a negative number.
So there are 4 things that can go down at this point.
1. We have everything positive [ex. log(3 x 10^3)].
2. We have n < 0 [ex. log(8.5 x 10^-15)].
3. We have the entire log negative [ex. -log(4 x 10^-7)].
4. We have only the log negative [ex. -log(1.95 x 10^6)].
It's important to know what happens with log(m) in all of these cases. Since the original expression contains 10 to some power, it's pointless to have m outside the range of [1, 10). Note that I do not include 10 in this range. If m were < 1 then you could multiply that by some power of 10 to make it within the range of [1, 10) and subsequently divide the exponential by the same power of 10 --> ex. 0.001 x 10^4. You can rewrite this by multiplying 0.001 by 1000 and dividing 10^4 by 1000 to get 1 x 10^1. If m were > or = 10 then you could divide that by some power of 10 to make it within the range of [1, 10) and subsequently multiply the exponential by the same power of 10 --> ex. 634.23 x 10^-14. You can divide 634.23 by 100 and multiply 10^-14 by 100 to get 6.3423 x 10^-12.
Anyhow, you should ALWAYS have m in that range. Since log(x) essentially answers the question '10 to what power equals x?' and since x is between 1 and 10, log(x) must be between 0 and 1 (not including 1). What happens when log(x) = 1/2? Well, log(x) = 1/2 means x = 10^(1/2) = sqrt(10) ~ 3.2. So log(3.2) ~ 1/2. That means 3.2 is a good judge of what log(m) will be. If m < 3.2 then log(m) < 1/2. If m > 3.2 then log(m) > 1/2.
Finally note how 3.2 is much less than halfway between 1 and 10 (midpoint = 5.5). Use that fact to judge approximate values. log(6.5) is about halfway between 3.2 and 10 so it's GREATER than 0.75 (halfway between 1/2 and 1) in the same way that log(5.5) is greater than 1/2 even though it's halfway between 1 and 10. In simpler terms, log(x) tends to increase slower as x increases.
Okay so...
1. All positive. We have log(m) + n. If m > 3.2 then the answer will be closer to n+1. If m < 3.2 then the answer will be closer to n.
Ex: log(7 * 10^8) = log(7) + 8 ~ 0.8 + 8 ~ 8.8. Actual answer 8.845.
2. Log positive, n negative. We have log(m) - n. If m > 3.2 then the answer will be closer to -n + 1. If m < 3.2 then the answer will be closer to -n.
Ex: log(3.2 * 10^-16) = log(3.2) - 16. Remember that log(3.2) ~ 0.5. So this is approximately 0.5 - 16 = -15.5. Actual answer -15.495.
3. All negative. So we start with -log(m x 10^-n) = -log(m) - log(10^-n) = -log(m) + nlog(10) = -log(m) + n. This is just case #2 * (-1). If m > 3.2 then the answer will be closer to n - 1. If m < 3.2 then the answer will be closer to n.
*NOTE that this is the most common case for pH problems!! That's why they always say the answer is between n-1 and n!!*
Ex: -log(2.45 * 10^-8) = -log(2.45) + 8 ~ -0.4 + 8 ~ 7.6. Actual answer 7.611.
4. Only log negative. This is just the negative of case #1. So we have -log(m) - n. If m > 3.2 then the answer will be closer to -n-1. If m < 3.2 then the answer will be closer to -n.
Ex: -log(9.8 * 10^10) = -log(9.8) - 10 ~ -1 - 10 ~ -11. Actual answer -10.991.
Hope this helps. I got bored from studying 🙂
Ah, you beat me to it, lol, very well said bro, you're my hero and kudos to you. Can I borrow your brain for the qr section, lol hope you're doing allright and we'll talk soon. Game day is on the 31st for me! Third time too, so there's a lot of pressure to really knock this one out of the park.
- Donjuan
