Question Number 10360 by Tawakalitu ayo mi last updated on 05/Feb/17

Solve for x in the equation.  3^x  + 4^x  + 5^x  = 6^x

Answered by mrW1 last updated on 05/Feb/17

when x=2,  3^2 +4^2 +5^2 =9+16+25=50  6^2 =36  i.e. 3^2 +4^2 +5^2 >6^2     when x=4,  3^4 +4^4 +5^4 =81+256+625=962  6^4 =1296  i.e. 3^4 +4^4 +5^4 <6^4     ⇒the solution is between 2 and 4.    through try and error we get the  solution x=3:  3^3 +4^3 +5^3 =27+64+125=216  6^3 =216

Commented bymrW1 last updated on 05/Feb/17

I′m not sure if there is an analytical  solution to this question.

Commented byTawakalitu ayo mi last updated on 05/Feb/17

God bless you sir.

Answered by arge last updated on 05/Feb/17

3^x +4^x =6^x −5^x     3^x =6^x   4^x = −5^x     ±4^x = −5^x   −4^x = −5^x     artificio,    3^x −4^x =6^x −5^x   x=0∵∵∵∵Rta

Commented byFilupSmith last updated on 06/Feb/17

if x=0, 3=1

Answered by chux last updated on 06/Feb/17

        i think it can be solved by linear approximation    (1+2)^x +(1+3)^x +(1+4)^x =(1+5)^x