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

2 Answers
Ratnaker Mehta
Mar 26, 2017

x=0.

Explanation:

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.!

Cesareo R.
Mar 26, 2017

See below.

Explanation:

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

Related questions
See all questions in Logarithm-- Inverse of an Exponential Function
Impact of this question
10019 views around the world
You can reuse this answer
Creative Commons License