Quadratic Equation containing exponential of variable

[Jayce]

Hi,

Can anyone help me in the proofing the value of R = 3.13 the following equation:

exp(-R/0.3) = 0.0003/R^2

 

[Deleted member 4993]

Hi,

Can anyone help me in the proofing the value of R = 3.13 the following equation:

exp(-R/0.3) = 0.0003/R^2

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Please share your work/thoughts about this assignment.

Do you know the Taylor polynomial expansion of e^-x?

These types are generally "approximated" through numerical method? Have you been taught such method/s?

 

[Jayce]

Please share your work/thoughts about this assignment.

I tried solving it using logarithm and exponential expansions as attached but no success.

With log expansion, I am getting answer 2.64 and using exponential expansion I am stuck at last steps in solving high degree equation.

 

[Steven G]

log(1+R-1) \(\displaystyle \neq\) log(1+R) - log(1)

 

[Steven G]

Your equation is a transcendental equation. Most transcendental equations can not be solved algebraically like you tried.

Is substitution enough for what you need to do.

 

[Jayce]

what range of R can be tried for substitution.

 

[JeffM]

Please share your work/thoughts about this assignment.

I tried solving it using logarithm and exponential expansions as attached but no success.

With log expansion, I am getting answer 2.64 and using exponential expansion I am stuck at last steps in solving high degree equation.

Like SK and Jomo, I suspect that this can be solved only approximately, and I am not sure what approximation techniques you know. What I am more concerned about are (1) you seem to have missed that you are asked to confirm an answer rather than find an answer, and (2) in your effort to find an answer you made basic errors.

[MATH]e^{(-r/0.3)} = \dfrac{3 * 10^{-4}}{r^2} = \dfrac{3}{10^4r^2}.[/MATH]
You now take logs of both sides without being consistent about the base of the logs being used.

[MATH] log_e(e^{(-r/0.3)} =log_e \left ( \dfrac{3}{10^4r^2} \right ) \implies[/MATH]
[MATH]-\ \dfrac{r}{0.3} = log_e(3) - 4log_e(10) - 2log_e(r) \ne log_e(3) - 4 - 2log_e(r).[/MATH]
You can't just change bases like that. There is a change of base formula. If you use the old-fasioned notation of ln for logs to the base of e and log without any base indicator for logs to the base of 10, you won't make that particular error again.

And then you made the really wild error of

[MATH]log(r) = log(1 + r - 1) = log(1 + r) - log(1) = log(1 + r) - 0 \implies log(r) = log(1 + r).[/MATH]
Oh my, oh my!

But apparently the question was

Is it true that

[MATH]e^{-(3.13/0.3)} \approx \dfrac{0.0003}{3.13^2}[/MATH]
How would you confirm or refute that?