Exponential Functions Homework: Proving at Least Two Solutions for 2^x = x^3

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Homework Help Overview

The discussion revolves around the equation 2^x = x^3, with participants tasked to demonstrate that this equation has at least two solutions. The subject area includes exponential functions and their intersections with polynomial functions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore different methods to identify solutions, including evaluating specific points, comparing the growth of the functions, and using logarithmic transformations. Questions arise regarding the validity of their approaches and the effectiveness of point selection for identifying zeros.

Discussion Status

There is an active exploration of various methods to find solutions, with some participants providing specific values and reasoning about intervals where solutions may exist. Guidance is offered regarding the continuity of functions and the implications of finding zeros, though no consensus on a single method has been reached.

Contextual Notes

Participants note the challenge of determining points where the two sides of the equation intersect and express uncertainty about the best strategies for identifying these points without resorting to guesswork.

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Homework Statement



2^x =x^3

show that the equation has at least two solutions.

Homework Equations


The Attempt at a Solution



I got xln2=3lnx but I don't know what I could do from here
 
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I took a differnet approach and thought about finding values where 2^x < x^3 then where 2^x > x^3

I found that between 1 and 3 there is a solution idk where the other one is though.
 
Try x=10.
 
i got f(2) = -4 f(0)=1 and it is continuos on this interval so there is a zero here and then

f(4) = -16
f(10) =24
and it is also continuous here so here is a zero

I subtracted the RHS so the equation is 2^x - x^3 =0

I just want to make sure this is a valid way to do this problem. And wondering if there is an easier way to pick points where I think a zero may occur other than guessing.
 
Well.. by the problem description, all you have to show is that there is 2 or more points which the equation is true, and if you can "literally" find the points that satisfies the equation, this definitely answers the question.
 
try this

2^x>x^3
you do logarithm operation on both of them

x* log 2 >3* log x (based 10)

x>[3/(log2)] * log x

now i think its easier to find this number by guessing
 

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