Writing Formula Where Using Natural logarithm Where Base is Unknown

Question

I want to rewrite a formula where the natural logarithm of a base is unknown. Specifically it is the revised formula for human to dog years, which is: $$\text{Human_Age} = 16 \times \ln(\text{Dog_Age}) + 31$$ This formula is written for when we know the dog ages, however I want rearrange this formula to where we known the human age and the dog age is unknown. Therefore I want a formula that converts human years to dog years, not the other way around.

This formula works for ages older than 1 for dog years, I've tried rearranging using Log but can't get it correct.

Answer 1

This formula looks ridiculous, and I recommend taking the results with a pinch of salt.

However, it can be rearranged as follows: $$\text{Dog_age}=\exp\left(\frac{\text{Human_age}-31}{16}\right);$$ here $\exp$ means the function $\exp(x)=e^x$, which is the inverse of $\ln$ (the formula was hard to read if I wrote it as a power).

If the original formula works for dog ages $\geq1$ then this works for human ages $\geq31$, since dog age $1$ (allegedly) corresponds to human age $31$.

Answer 2

First, the outermost function is addition of 31, and the inverse of addition is subtraction. Thus, $$\text{Human_Age}-31 = 16 \times \ln(\text{Dog_Age}).$$

Second, then next outermost function is multiplication by 16, and the inverse of multiplication is division. Thus, $$\frac{\text{Human_Age}-31}{16} = \ln(\text{Dog_Age}).$$

Third, the next (and last) outermost function is the natural logarithm, and the inverse of the natural logarithm is the natural exponentiation (that is, $e^x$). Thus, $$e^{\frac{\text{Human_Age}-31}{16}} = \text{Dog_Age}. $$

Or $$\text{Dog_Age} = e^{\frac{\text{Human_Age}-31}{16}}.$$

The key is that the outermost function in the direct direction becomes the innermost function in the inverse direction, and vice versa.