The-number-of-real-solutions-of-3-x-4-x-5-x-is-

Question Number 98057 by PengagumRahasiamu last updated on 11/Jun/20

The number of real solutions of

3^{x} + 4^{x} = 5^{x} is____.

Answered by Rio Michael last updated on 11/Jun/20

let f (x) = 3^{x} + 4^{x} - 5^{x} = 0

for x ⩽ 2, f (x) is positive .

for x x ⩾ 3, f (x) is negative .

so f (x) is a decreasing function thus

it must have only one root between 2 and 3 .

x \approx 2.73 \dots

Commented by mr W last updated on 11/Jun/20

x = 2 :

3^{2} + 4^{2} = 5^{2}

Answered by 1549442205 last updated on 12/Jun/20

We prove that this equation has unique

real root x = 2 . Indeed, dividing both two sides of

given equation by 5^{x} we get

{(\frac{3}{5})}^{x} + {(\frac{5}{5})}^{x} = 1 . It is easy to see that x = 2 satisfy since {(\frac{3}{5})}^{2} + {(\frac{5}{5})}^{2} = \frac{9}{25} + \frac{16}{25} = 1

which means that x = 2 be a root of given eq .

Also, it is easy to see that the functions

f (x) = {(\frac{3}{5})}^{x} and g (x) = {(\frac{4}{5})}^{x} are decreasing on (- \infty; + \infty)

because f^{'} (x) = {(\frac{3}{5})}^{x} \ln \frac{3}{5} < 0, g^{'} (x) = {(\frac{4}{5})}^{x} \ln \frac{4}{5} < 0

on (- \infty; + \infty) . Hence,

a / For x < 2 we have {(\frac{3}{5})}^{x} + {(\frac{5}{5})}^{x} > {(\frac{3}{5})}^{2} + {(\frac{5}{5})}^{2} = 1

b / For x > 2 we have {(\frac{3}{5})}^{x} + {(\frac{5}{5})}^{x} < {(\frac{3}{5})}^{2} + {(\frac{5}{5})}^{2} = 1

That show that x = 2 is unique real root of

the given equation