Is it Possible to solve Exponential Equations like these?

AI Thread Summary
Exponential equations like 8000 - 1.2031 * e^(0.763x) = 0.5992 * e^(0.7895x) are typically transcendental and cannot be solved algebraically. Numerical or graphical methods are recommended for finding solutions, with iteration using trial values of x being effective. One participant noted that the second exponential is approximately half of the first, suggesting x is around 10. They also proposed rewriting the equation to simplify the terms for a more accurate solution. Overall, algebraic solutions are not feasible, and numerical approaches are necessary for resolution.
Alex Myhill
Messages
5
Reaction score
0

Homework Statement


Hi, I have come across this equation in modelling exponential growth and decay. I am wondering if it is possible to solve it algebraically or not?

Homework Equations


8000-1.2031 * e^ (0.763x)=(0.5992×e^0.7895x)

The Attempt at a Solution


Brought all e^(ax) values to one side, however beyond that step I am really not sure of where to go, have looked at the problem for several hours without any results, any help would be appreciated.
 
Physics news on Phys.org
Alex Myhill said:

Homework Statement


Hi, I have come across this equation in modelling exponential growth and decay. I am wondering if it is possible to solve it algebraically or not?

Homework Equations


8000-1.2031 * e^ (0.763x)=(0.5992×e^0.7895x)

The Attempt at a Solution


Brought all e^(ax) values to one side, however beyond that step I am really not sure of where to go, have looked at the problem for several hours without any results, any help would be appreciated.
In general, there is no algebraic method to solve such equations since they are transcendental rather than algebraic by nature.

Only a numerical or graphical solution can be obtained. Iteration using different trial values of x is probably the quickest way to find a solution here.
 
Hi SteamKing, thankyou for your answer, I have learned something from that.
 
In this particular equation, you can easily see that the second exponential expression is likely to be about half the first, so x must be about 10.
You could then rewrite the .7895x as .763x+.0265x, or approximately .763x+0.265. That should get you to a reasonably accurate answer. You could redo that with the more accurate x value as a check.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top