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#### PEPAKURAPROP1

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##### How would Q10 equations be solved?
« on: May 17, 2017, 09:19:45 PM »

Hi, I'm studying this lesson in my class, but I have absolutely no understanding how my instructor has done this equation. I have been looking this up, but it's making absolutely no sense to me. We are doing Q10 equations. These equations are suppose to estimating the shelf life of pharmaceutical drugs. These are the 6 answers

T 90 (T 2 )= T 90 (T 1 )/Q 10 (ΔT/10)

Example of how instructor does it-

Calculate the factors by w/c rate constant my change for
A) A 25°c to 50°c temperature change
b) 25°c to 0°c change
Q 10 = 2, 3, 4

Q 10 = 2
ΔT=  (50 - 25)= 25
2 [/sub]25/10[/sup] = log2 [/sup]2.5[/sup]
Antilog= .7525
= 5.7

Q 10 = 3
3 25/10 = log 3 2.5 = .4771(2.5)
Antilog= 1.1928
=15.6

Q 10 = 4
4 25/10 =  log4 2.5 = .6021 (2.5)
Antilog= 1.505
= 32

ΔT= (T 2
= (0 - 25)
= - 25
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Q 10 = 2

=1/log2 2.5 = 1/.301(2.5)=1/.7525=1/5.9
~~~~
Q 10 = 3

=1/log3 2.5 =1/0.4771(2.5)=1/antilog1.1928=1/15.6
~~~~
Q 10 =4

1/log4 2.5 = 1/.904(2.5)= 1/Antilog 1.505 = 1/32

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1. How did she get the log and antilog?
If there are small mistakes, please forgive me, I'm new to this site, sorry.
« Last Edit: May 17, 2017, 09:32:43 PM by Borek »
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#### Borek

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##### Re: How would Q10 equations be solved?
« Reply #1 on: May 17, 2017, 09:45:33 PM »

This part looks formatted well enough to understand what is going on:

3 25/10 = log 3 2.5 = .4771(2.5)
Antilog= 1.1928
=15.6

This is not exactly true, equal sign is misused. Looks like we are trying to calculate the value of the expression

$$3^{\frac{25}{10}}$$

This can be indeed make easier with a use of logarithms (assuming you know their properties):

$$log(3^{\frac{25}{10}}) = \frac{25}{10}log(3)$$

log(3) can be taken from tables, no other simple way of getting its value (well, you can use calculator instead of tables, but then it will be typically much easier to just directly calculate $3^{\frac{25}{10}}$ without additional tricks).

So

$$log(3^{\frac{25}{10}}) = \frac{25}{10}log(3) = 2.5\times0.477121 = 1.1928$$

and finally, using log properties again

$$3^{\frac{25}{10}} = 10^{1.1928} = 15.5885$$

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#### PEPAKURAPROP1

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##### Re: How would Q10 equations be solved?
« Reply #2 on: May 18, 2017, 09:09:36 PM »

Hello, and thank you for your reply, I appreciate you spending time to reply to my post, and thank you! Only one part I don't get though, the bottom of the equation, how it simplified?

3* 25/10 = 101.1928 = 15.5885
« Last Edit: May 18, 2017, 09:24:42 PM by PEPAKURAPROP1 »
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