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

Mar 26, 2017

$x = 0.$

#### Explanation:

Let ${2}^{x} = a , {3}^{x} = b , \mathmr{and} , {5}^{x} = c .$

$\therefore {4}^{x} + {9}^{x} + {25}^{x} = {a}^{2} + {b}^{2} + {c}^{2.} \ldots \ldots \ldots \ldots . \left(1\right) .$

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,

${a}^{2} + {b}^{2} + {c}^{2} - a b - b c - c a = 0 , \mathmr{and} ,$

$\frac{1}{2} \left\{{\left(a - b\right)}^{2} + {\left(b - c\right)}^{2} + {\left(c - a\right)}^{2}\right\} = 0.$

$\Rightarrow a = b = c , \mathmr{and} , {2}^{x} = {3}^{x} = {5}^{x} .$

${2}^{x} = {3}^{x} \Rightarrow {\left(\frac{2}{3}\right)}^{x} = 1 = {\left(\frac{2}{3}\right)}^{0}$

$x = 0.$

${3}^{x} = {5}^{x} \text{ also } \Rightarrow x = 0.$

$\therefore x = 0$ is the Soln.

Enjoy Maths.!

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

$u = \left(X , Y , Z\right) , v = \left(X , Z , Y\right) , w = \left(Y , X , Z\right)$

we have

$\left\langlev , w\right\rangle = \left\lVert v \right\rVert \left\lVert w \right\rVert \cos \left(\angle \left(v , w\right)\right)$

where $\left\langle\cdot , \cdot\right\rangle$ represents the scalar product of two vectors.

with $\left\mid \cos \left(\angle \left(v , w\right)\right) \right\mid \le 1$

but $\left\lVert u \right\rVert = \left\lVert v \right\rVert = \left\lVert w \right\rVert$

so

$\left\langleu , u\right\rangle = {\left\lVert u \right\rVert}^{2} \ge \left\langlev , w\right\rangle$

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

$X = Y = Z \to {2}^{2} = {3}^{x} = {5}^{x} \to x = 0$