exponential functions

[physstudent1]

1. The problem statement, all variables and given/known data

2^x =x^3

show that the equation has at least two solutions.

2. Relevant equations



3. The attempt at a solution

I got xln2=3lnx but I don't know what I could do from here

[physstudent1]

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.

[morphism]

Try x=10.

[physstudent1]

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 continous 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.

[l46kok]

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.

[transgalactic]

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