Question

solutions of 2^x +3^x +4^x -5^x=0???

Answer 1

Are you misreading, Anusha? The equation is:
\[2^x+3^x+4^x-5^x=0\]

Answer 2

Consider numerical testing. Test the equation at \(x=1\). Now test it at \(x=2\). Then try \(x=3\). See the idea?

Answer 3

log(0) does not exist; so logging both sides is futile

Answer 4

Do you want integral solutions?

Answer 5

and you'll require the knowledge of high degree interpolation i guess in this type of ques..

Answer 6

Oh my good lord, this is the root of a ridiculous exponential equation.

Answer 7

You will definitely need interpolation.

Answer 8

ya the solution can be of any type it can be integral or complex

Answer 9

interpolation???

Answer 10

Interpolation is how you find the approximation of a solution to an equation by narrowing the interval of potential solutions over and over.

Answer 11

hmm thanks for the help:)

Answer 12

i always called it "trial and error"

Answer 13

Hmmmm.
@Limitless What if I take all the terms(2^x, 3^x....) in terms of a common factor and then solve it.?

Answer 14

It is definitely not looking pretty in any way what-so-ever. I don't think it was intended to be solved in an analytical way. Maybe I am wrong.

Answer 15

i see one place we can simplify

2^x +3^x +4^x -5^x=0 ; 4^x = 2^(2x)

2^x +3^x + 2^(2x) -5^x=0

but im not sure of thats useful or not :/

Answer 16

id say give it a shot, the worst that could happen is it doesnt pan out; and even if its not the right method, you might even stumble upon a right answer :)

Answer 17

its all in the dbl chk

Answer 18

Okay. Damn. I'm sorry. It doesnt turn into a linear equation. My bad.

Answer 19

2^x +3^x +4^x -5^x=0

2^x +2^(2x) = 5^x - 3^x

2^x( 1 + 2^x) = 5^x - 3^x

1 + 2^x = (5^x - 3^x)/2^x

http://www.wolframalpha.com/input/?i=2%5Ex+%2B3%5Ex+%2B4%5Ex+-5%5Ex%3D0

ewww ....

Answer 20

Well.. There is an exponential equation of which it is a root. I thought you might be onto something, but I could not tell.

Answer 21

can we turn it into a taylor and give it a shot?

Answer 22

I'm not sure how we'd convert. I tried using the binomial theorem with every term in terms of \(2^x\). It does not look pretty nor useful.

Also, I can't remember if there is an \(n\)th derivative of the general exponential function..

Answer 23

a^x (ln(a))^n i think

Answer 24

Then there is certainly a taylor series. It would be simply
\[\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n=0\]
where we set \(f(x)=2^x+3^x+4^x-5^x\).
I am not entirely sure what would be the best choice of \(a\).

Answer 25

a = 0 or 1 is my idea; but im sure there is an interval of convergence to consider

Answer 26

I don't know what the interval of convergence is for a general taylor series off of the top of my head. Would it be just like the radius of convergence for a power series?

Answer 27

yes, since a taylor is just a powr series

Answer 28

ln2 + ln3 + ln4 - ln5 = ln(2.3.4/5) = ln(24/5) if we pick a=0

Answer 29

I'm trying to think we need to consider the interval of convergence first. I don't know how, though.. >_<

Answer 30

The fact that \(a\) is part of the coefficient has totally thrown me off.

Answer 31

\[T(x) = 2 + \sum_{n=1}^{inf}\frac{ln^n(24/5)x^n}{n!}\]

\[2+\lim\frac{ln^n(24/5)x^n}{n!}*\frac{(n-1)!}{ln^{(n-1)}(24/5)x^{(n-1)}}\]
\[2+\lim\frac{ln(24/5)x}{n}\]
\[2+|ln(24/5)x|\lim\frac{1}{n}\]
as n to inf; thats 0

and 0<1 always so the interval might be everywhere convergent ..... just a guess tho

Answer 32

http://www.wolframalpha.com/input/?i=sum%28n+from+1+to+inf%29+%28ln%2824%2F5%29%29%5Enx%5En%2Fn%21+and+2%5Ex%2B3%5Ex%2B4%5Ex-5%5Ex-2

its not everywhere convergent sooo, i missed it in the figuring :)

Answer 33

It seems to be badly approximating our function, too...

http://www.wolframalpha.com/input/?i=sum%28n+from+1+to+inf%29+%28ln%2824%2F5%29%29%5Enx%5En%2Fn%21+and+2%5Ex%2B3%5Ex%2B4%5Ex-5%5Ex