Ratio of a/b when lna/lnb=x

AlexRaw

New member
Hi,
I am working on a data set and I have log-transformed two variables a and b and would like to figure out the relationship between the respective ratios. I have a strong intuition that a/b should have a well-defined range for a given value of lna/lnb, but my maths is poor.

a and be are positive integers with a finite range
0 < a < amax, amax=499639
1 < b < bmax, bmax=837481

I am looking for the range of
a/b
when
ln(a+1)/ln(b+1) = x
0 < x < 1

I know that for x = 1, a/b = 1 and when x = 0, a/b = 0
For anything in between there must be a range.

I'd like to plug in an x and get a range of possible values for a/b.
I have run a regression analysis on my data and figured out the deviations, but I am looking for a cleaner, more general solution that I could potentially also compute when the ranges of a and b are different.

Is this even possible?

Thanks in advance. I'd appreciate any help.

Llwewllyn

New member
*Note, it usually helps to write out the mathematical computations yourself for better understanding. Also, the final solution that I have concluded will be capitalized and spaced out to make the conclusion clearer and stand out.*

Hi, I understand that it has been quite some time since you posted this question, but I would like to answer it simply for the sake of not leaving this question unanswered.

Now I'm not sure if there is something missing to this question, but as far as I understand it:

0 < a < 499,639

1 < b < 837,481; where:

ln(a+1)/ln(b+1) = x and 0 < x < 1

Understanding the premise of the question, lets simplify the information given into something more palatable:

ln(a+1)/ln(b+1) = x

ln((a+1) - (b+1)) = x

ln(a+1-b-1) = x

ln(a-b) = x

(a-b) = e^x; where 0 < x < 1

Now that we have simplified, we have two pieces of information:

(a-b) = e^x; where 0 < x < 1

a/b = x

With this in mind, we can combine these equations to further simplify this into one equation:

( A - B ) = E ^ ( A / B ); where 0 < a/b < 1, 0 < a < 499,639 and 1 < b < 837,481

This equation can be graphed using the appropriate restrictions in order to achieve the graph you (possibly) are looking for. I hope this helps!

* the solution that satisfies the given restrictions are where the green and purple colors MEET, that is, the darker purple color houses all of the possible combinations of "a" and "b" (x and y) that satisfies the conditions outlined in the question.*

mrtwhs

Active member
ln(a+1)/ln(b+1) = x

ln((a+1) - (b+1)) = x

ln(a+1-b-1) = x

ln(a-b) = x
$$\displaystyle \ln(m-n) = \ln \left( \dfrac{m}{n} \right) \neq \dfrac{\ln m}{\ln n}$$

topsquark

Well-known member
MHB Math Helper
$$\displaystyle \ln(m-n) = \ln \left( \dfrac{m}{n} \right) \neq \dfrac{\ln m}{\ln n}$$
Sorry, but $$\displaystyle ln(m - n) \neq ln \left ( \dfrac{m}{n} \right )$$.

$$\displaystyle ln \left ( \dfrac{m}{n} \right ) = ln(m) - ln(n)$$

@Llwewllyn:
Neither $$\displaystyle ln \left ( \dfrac{m}{n} \right)$$ nor $$\displaystyle \dfrac{ln(m)}{ln(n)}$$ are equal to $$\displaystyle ln(m - n)$$. Unless we are using some kind of approximation there is no way to rewrite $$\displaystyle ln(m - n)$$ in general.

-Dan

mrtwhs

Active member
Sorry, but $$\displaystyle ln(m - n) \neq ln \left ( \dfrac{m}{n} \right )$$.

$$\displaystyle ln \left ( \dfrac{m}{n} \right ) = ln(m) - ln(n)$$

@Llwewllyn:
Neither $$\displaystyle ln \left ( \dfrac{m}{n} \right)$$ nor $$\displaystyle \dfrac{ln(m)}{ln(n)}$$ are equal to $$\displaystyle ln(m - n)$$. Unless we are using some kind of approximation there is no way to rewrite $$\displaystyle ln(m - n)$$ in general.

-Dan
Send me to the woodshed! I misplaced my parentheses.

Olinguito

Well-known member
$$\displaystyle \ln(m-n) = \ln \left( \dfrac{m}{n} \right) \neq \dfrac{\ln m}{\ln n}$$
You mean
$$\displaystyle \ln(m-n) \neq \ln \left( \dfrac{m}{n} \right) = \dfrac{\ln m}{\ln n}$$

topsquark

Well-known member
MHB Math Helper
You mean
$$\displaystyle \ln(m-n) \neq \ln \left( \dfrac{m}{n} \right) = \dfrac{\ln m}{\ln n}$$
Or rather $$\displaystyle \ln(m-n) \neq \ln \left( \dfrac{m}{n} \right) \neq \dfrac{\ln m}{\ln n}$$

-Dan