Logarithms and exponents

  • Thread starter Intr3pid
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  • #1
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hello everyone

I'm really stuck on logs. My teacher doesn't take pride in her work and teaches very sloppy stuff. I was wondering if I can get help with these questions.

1) log to the base 3 of x - log to the base 6 of x = 2

2) a lab tecnician places a bacterial cell into a vial at 5 am. The cells divide in such a way that the number of cells doubles every 4 minutes. The vial is full 1 hour later.

a) how long does it take fot the cells to divide to 4095?

b) at what time is the vial full?

c) at what time is the vial 1/16 full?

for #2, how do you know which "formula" to use. My teacher presented me with 3 formulas:

Ptotal = Pone x e^kt
Atotal = Aone x C^t/p
A = Pe^it

e= is the irrational number 2.71828...

thanks in advance
 

Answers and Replies

  • #2
3,763
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What have you done so far ?

Here are some general rules

[tex]log_a(AB) = log_a(A) + log_a(B)[/tex]
[tex]log_a(A/B) = log_a(A) - log_a(B)[/tex]
[tex]log_a(A^n) = nlog_a(A) [/tex]
To change base :
[tex]log_a(A) = \frac{log_b(A)}{log_b(a)}[/tex]

and finally
[tex]ln(x) = log_e(x)[/tex]
[tex]log_a(x) = B <--> a^B = x[/tex]

marlon

EDIT : Here is some more
 
Last edited:
  • #3
HallsofIvy
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Intr3pid said:
hello everyone
I'm really stuck on logs. My teacher doesn't take pride in her work and teaches very sloppy stuff. I was wondering if I can get help with these questions.
1) log to the base 3 of x - log to the base 6 of x = 2
When dealing with logarithms to different bases it is best to remember the formula
[tex]log_a(x)= \frac{log x}{log a}[/itex]
where the logarithms on the right side can be to any base.
In your problem
[tex]log_3 x- log_6 x= \frac{log x}{log 3}- \frac{log x}{log 6}= 2[/tex]
add those fractions and then use some of the "laws of logarithms" marlon listed.
2) a lab tecnician places a bacterial cell into a vial at 5 am. The cells divide in such a way that the number of cells doubles every 4 minutes. The vial is full 1 hour later.
a) how long does it take fot the cells to divide to 4095?
b) at what time is the vial full?
c) at what time is the vial 1/16 full?
for #2, how do you know which "formula" to use. My teacher presented me with 3 formulas:
Ptotal = Pone x e^kt
Atotal = Aone x C^t/p
A = Pe^it
e= is the irrational number 2.71828...
thanks in advance
Actually those formulas are basically the same formula in different guises.
The third is exactly the same as the first with A= Ptotal, P= Pone,and i= k.
The second is more subtle but is basically the same.
One way to do this problem using the first or third equations: When t= 0, ek0= e00 so Ptotal= Pone= 1 since there was one bacterium when t= 0. After 4 minutes, that bacterium has doubled to 2 bacteria: as long as t is measured in minutes, Ptotal= 2= 1(ek*4) so e4k= 2. Taking ln of both sides k= (1/4)ln(2). Now you know that Ptotal= e(ln 2)(t/4). To answer (a), solve the equation
e(ln 2)t= 4095 for t.
Another way, using the second equation above: Since the number of bacteria double every 4 minutes, we could just multiply by t every 4 minutes: if t is measured in minutes, that means we double every t/4 so
Atotal= Aone (2)t/4. Again, if t= 0, Atotal= 1= Aone so the formula is just Atotal= 2t/4. Solve 2t/4= 4095 to t. You can get a good approximation to that by recognizing that 4096 (4K to our computer friends) is 212.
Those two formulas: Atotal= 2t/4 and Ptotal= e(ln 2)(t/4) are exactly the same because, using the formulas marlon gave, (ln 2)t= ln(2t) and [itex]e^{ln(2^t}= 2^t.

b) at what time is the vial full?
Well, you were told one hour (60 minutes) weren't you?

c) at what time is the vial 1/16 full?
jWork backwards from one hour. Since the number of bacteria double every 4 minutes, 4 minutes before the hour (56 minutes) the vial must have been 1/2 full. 4 minutes before that (52 minutes) the vial must have been 1/4 full, etc.
 
  • #4
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HallsofIvy said:
When dealing with logarithms to different bases it is best to remember the formula
[tex]log_a(x)= \frac{log x}{log a}[/itex]
where the logarithms on the right side can be to any base.
In your problem
[tex]log_3 x- log_6 x= \frac{log x}{log 3}- \frac{log x}{log 6}= 2[/tex]
add those fractions and then use some of the "laws of logarithms" marlon listed.
Actually those formulas are basically the same formula in different guises.
The third is exactly the same as the first with A= Ptotal, P= Pone,and i= k.
The second is more subtle but is basically the same.
One way to do this problem using the first or third equations: When t= 0, ek0= e00 so Ptotal= Pone= 1 since there was one bacterium when t= 0. After 4 minutes, that bacterium has doubled to 2 bacteria: as long as t is measured in minutes, Ptotal= 2= 1(ek*4) so e4k= 2. Taking ln of both sides k= (1/4)ln(2). Now you know that Ptotal= e(ln 2)(t/4). To answer (a), solve the equation
e(ln 2)t= 4095 for t.
Another way, using the second equation above: Since the number of bacteria double every 4 minutes, we could just multiply by t every 4 minutes: if t is measured in minutes, that means we double every t/4 so
Atotal= Aone (2)t/4. Again, if t= 0, Atotal= 1= Aone so the formula is just Atotal= 2t/4. Solve 2t/4= 4095 to t. You can get a good approximation to that by recognizing that 4096 (4K to our computer friends) is 212.
Those two formulas: Atotal= 2t/4 and Ptotal= e(ln 2)(t/4) are exactly the same because, using the formulas marlon gave, (ln 2)t= ln(2t) and [itex]e^{ln(2^t}= 2^t.
b) at what time is the vial full?
Well, you were told one hour (60 minutes) weren't you?
c) at what time is the vial 1/16 full?
jWork backwards from one hour. Since the number of bacteria double every 4 minutes, 4 minutes before the hour (56 minutes) the vial must have been 1/2 full. 4 minutes before that (52 minutes) the vial must have been 1/4 full, etc.

i was wondering how u add log fractions?
 
  • #5
Ouabache
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Intr3pid said:
i was wondering how u add log fractions?
If you mean fractions like this the one HallofIvy described for your part (1),
just factor out log(x) from each fraction, what you have left after evaluating the logs in denominators) are two contants which can be subtracted. The right side is still a value of 2. You can now solve the resulting equation algebraically for log(x).
 
  • #6
HallsofIvy
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The same way you do any fractions!
[tex]log_3 x- log_6 x= \frac{log x}{log 3}- \frac{log x}{log 6}= 2[/tex]
[tex](log x)\frac{1}{log 3}-\frac{1}log 6}= 2[/tex]
[tex](log x)\frac{log 6- log 3}{(log 6)(log 3)}= 2[/tex]
[tex](log x)\frac{log 2}{(log 6)(log 3)}= 2[/tex]
[tex]log x= \frac{2(log 6)(log 3)}{log 2}[/itex]
Use your calculator to find the right hand side, then solve for x.
 
  • #7
Ouabache
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That's what I said. But after the 2nd line that HallsofIvy derived,
I algebraically rearranged terms..
logx = 2/[(1/log3) - (1/log6)]
and evaluated the right handside with a calculator.
If you are asked to solve for x, just take the inverse log.
 
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