Question

Find the Number of solutions

Answer 1

\[2^x +3^x + 4^x- 5^x=0\]

Answer 2

Hard to imagine there being more than one solution unless x can be complex.

Answer 3

what are the conditions, is x supposed to be a positive integer?

Answer 4

Real values only.

Answer 5

If x is Real, we can show there is one solution between 2 and 3.

Answer 6

@Valpey How?

Answer 7

https://www.wolframalpha.com/input/?i=y+%3D+2%5Ex%2B3%5Ex%2B4%5Ex%E2%88%925%5Ex+for+x+from+0+to+3

Answer 8

http://www.wolframalpha.com/input/?i=solve+2^x+%2B3^x+%2B+4^x-+5^x%3D0

Answer 9

you can show there is a solution using intermediate value theorem

Answer 10

and using calculus

Answer 11

Yea i tried that.
But can you tell me a more theoretical way to show there is a real number b/w 2 and 3 which the solution?

Answer 12

Haven't Studied it yet. :/
But i know something about it as learned fromm the net.

Answer 13

oh ohk
got it thanks.
but do we have to prove its a continuous curve or is it obvious?

Answer 14

let f(x) =2^x +3^x + 4^x- 5^x
f(2) > 0 (above the x axis)

f(3) < 0 (below the x axis)

because of continuity it must cross or intersect the x axis .
we formally state this using the 'intermediate value theorem'

Answer 15

you can say, since each exponential function by itself is continuous on the reals, and the sum of continuous functions is continuous, that expression is continuous

Answer 16

also you can plug in negative infinity and positive infinity to establish the end behavior of the function .
as x goes to negative infinity the y value goes to zero asymptotically.
as x goes to positive infinity the y value drops to negative infinity.