Question

plzzz answer no. ii in the following image through indices

Answer 1

method 1 :-
=================={
(x√x )^x = x^(2x√x)

{ x¹.x½ }^x = x^{ 2x¹.x½ }

{x^3/2}^x = x^{ 2.x^3/2 }

take log both sides,
log{ x^3/2 }^x = logx^{2.x^3/2}

x.log{x^3/2} =2.x^3/2logx

[ logx^m = mlogx use here, ]

3/2. x logx = 2.x^3/2 logx

3x/2 logx - 2.x^3/2logx = 0

logx { 3x/2 - 2x^3/2} = 0

logx = 0 and { 3x/2 - 2x^3/2 } = 0

logx = 0
x = 1

again,

{3x/2 - 2x^3/2} = 0

3x/2 = 2x^3/2

take square both sides,

9x²/4 = 4x³

9x² = 16x³

x²( 16x - 9) = 0

x = 0 and x = 9/16

but x ≠ 0 becoz for log x > 0

hence,
x = 1 and 9/16

method 2 :-
==================

(x√x)^x = x^(2x√x)

(x¹.x½)^x = x^(2x¹.x½)

(x^3/2)^x = x^(2x^3/2 )

x^(3x/2) = x^(2x^3/2 )

[ x^m = x^n then , m = n use here ]

3x/2 = 2x^3/2

take square both sides,

9x²/4 = 4x³

x²( 9 - 16x) = 0

x = 0, and 9/16
but we know according to exponential coencept a^x then, a >0
hence,
x ≠ 0
so, x = 9/16
but also if we check at x = 1

LHS = (x√x )^x = 1¹ = 1
RHS = x^(2x√x) = 1² = 1

hence, LHS = RHS
so, answer is x = 1 and 9/16