How to solve this?#4^x+9^x+25^x=6^x+10^x+15^x#

Question

9692 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-26T11:04:53

#x=0.#

Let #2^x=a, 3^x=b, and, 5^x=c.#

#:. 4^x+9^x+25^x=a^2+b^2+c^2..............(1).#

Also, #6^x={(2)(3)}^x=2^x3^x=ab, &, ||ly, 10^x=ca, 15^x=bc...(2)#

Hence, using #(1), & (2),#the given eqn. becomes,

#a^2+b^2+c^2-ab-bc-ca=0, or,#

#1/2{(a-b)^2+(b-c)^2+(c-a)^2}=0.#

#rArr a=b=c, or, 2^x=3^x=5^x.#

#2^x=3^x rArr (2/3)^x=1=(2/3)^0#

#x=0.#

#3^x=5^x" also "rArr x=0.#

#:. x=0# is the Soln.

Enjoy Maths.!

Answer 2 · 2017-03-26T11:20:08

See below.

Calling #X=2^x, Y=3^x,Z=5^5# we have

#X^2+Y^2+Z^2=X Y+X Z+Y Z#

Now calling

#u = (X,Y,Z), v=(X,Z,Y), w = (Y,X,Z)#

we have

#<< v, w >> = normv norm w cos(angle(v,w))#

where #<< cdot, cdot >># represents the scalar product of two vectors.

with #abs(cos(angle(v,w))) le 1#

but #norm u=norm v=norm w#

so

#<< u, u>> = norm u^2 ge << v, w >> #

and the equality is attained only if #u = v = w# so

#X=Y=Z->2^2=3^x=5^x->x=0#