Can You Solve the Equation 5^n = 80n for n?

  • Context: High School 
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Discussion Overview

The discussion revolves around solving the equation 5^n = 80n for the variable n. Participants explore potential solutions, including both integer and real number solutions, and share their findings and methods.

Discussion Character

  • Homework-related, Mathematical reasoning, Exploratory

Main Points Raised

  • One participant expresses difficulty in solving the equation and seeks assistance.
  • Another participant claims that there are no integer solutions for n.
  • A participant reformulates the equation as 5^x - 80*x = 0 and suggests plotting it, reporting two approximate solutions: n ≈ 0.012759 and n ≈ 3.50133.
  • Further, two exact solutions in terms of the ProductLog function are provided, with detailed expressions and numerical approximations to high precision.
  • A participant questions whether the provided solutions can be considered 'analytical'.
  • A participant thanks others for their replies.

Areas of Agreement / Disagreement

Participants generally agree that there are no integer solutions, but multiple competing views exist regarding the nature and existence of real solutions, with some participants providing specific values and expressions.

Contextual Notes

The discussion includes assumptions about the types of solutions (integer vs. real) and relies on the definitions of the ProductLog function, which may not be universally understood. The mathematical steps leading to the exact solutions are not fully resolved.

One-D
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I've got a problem with this:
5^n=80n
find n!
I can't solve out this one.
thanx for ur help, I'm not sure if it's the right place for this thread.
 
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If [tex]n \in \mathbb{Z}[/tex], there is no such number.
 
5^x - 80*x = 0
plot it
and so far I've found 2 solutions (oh and there are no solns in Z of course)

Approximate roots are:

n=0.012759
n=3.50133
 
Last edited:
Anyway the 2 exact solutions (in real numbers) are:

[tex]n = -\frac{\text{ProductLog} \left( - \frac{\ln 5}{80} \right)}{\ln 5}[/tex]

and

[tex]n = -\frac{\text{ProductLog} \left(-1, - \frac{\ln 5}{80} \right)}{\ln 5}[/tex]

Which works out to 100 sf.:

n = 0.01275934595849510712320293028417381446823346765193827957982622843955432250672941443435143426214944153

And:

n = 3.501327193786496727104387840526356291207393167955463046140101380079871277483867981890189846055191784
 
and that is an 'analytical' solution?
 
thanx for ur replies
 

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