Answer:
[tex] x =2[/tex]
Step-by-step explanation:
Given :-
- [tex]{3}^{x} + {4}^{x} = {5}^{x} [/tex]
And we need to find out the value of x. Well there is no specific method to solve the equation .This can be only done using the " Trial and error" Method.
- We know that , 3 , 4 and 5 are Pythagorean triplets . So the sum of squares of two smallest numbers is equal to the square of the largest number . Henceforth ,
[tex]\implies {3}^{2} + {4}^{2} = {5}^{2} [/tex]
Verification :-
[tex]\implies {3}^{2} + 4^2 = 9+16=25=\boxed{5^2}[/tex]
So , the value of x is 2 . We can here prove that , x does not have other roots other than 2 . For that , divide the both sides of equation by [tex]5^x[/tex] , we have ,
[tex]\implies \dfrac{{3}^{x} + {4}^{x}}{5^x} = \dfrac{{5}^{x}}{5^x} [/tex]
[tex]\implies \bigg( \dfrac{3}{5}\bigg)^x+ \bigg( \dfrac{4}{5}\bigg)^x = 1 [/tex]
- Now if we take the value of x greater than 2 or less than 2 , then the value 1 will not be satisfied for the values of x greater than or less than 2 .
That is ,
[tex]\implies \bigg( \dfrac{3}{5}\bigg)^x+\bigg( \dfrac{4}{5}\bigg)^x > \bigg( \dfrac{3}{5}\bigg)^2 +\bigg( \dfrac{4}{5}\bigg)^2 [/tex]
Subsequently :-
[tex]\implies \bigg( \dfrac{3}{5}\bigg)^x+\bigg( \dfrac{4}{5}\bigg)^x > 1 [/tex]
[tex]\implies \bigg( \dfrac{3}{5}\bigg)^x+\bigg( \dfrac{4}{5}\bigg)^x < \bigg( \dfrac{3}{5}\bigg)^2 +\bigg( \dfrac{4}{5}\bigg)^2 [/tex]
Subsequently :-
[tex]\implies \bigg( \dfrac{3}{5}\bigg)^x+\bigg( \dfrac{4}{5}\bigg)^x <1[/tex]
- Thus there is no other value other than 2 for which the value of above expression becomes 1 .
Hence 2 is the root of the given equation.
