Question

if x^m = y^n and y=x^2,prove that m-2n = 0​

Answer 1

Answer:

$Given: xmyn=(x+y)m+n</p><p></p><p>Take log on both side</p><p></p><p></p><p>logxm+logyn+(m+n)log=y(x+y)</p><p></p><p>mlogx+nlogy=(m+n)log(x+y)</p><p></p><p></p><p>Differentiating w.r.t x on both sides we get,</p><p></p><p></p><p>m.x1+ny1dxdy=(m+n).(x+y)1(1+dxdy)</p><p></p><p>xm+yndxdy=x+ym+n(1+dxdy)</p><p></p><p>dxdy(yn−x+ym+n)=x+ym+n−xm</p><p></p><p>dxdy(y(x+y)n(x+y)−y(m+n))=x(x+y)x(m+n)−m(x+y)</p><p></p><p></p><p>On simplifying we get,</p><p></p><p></p><p>dxdy(ynx−my)=xnx−my</p><p></p><p>∴dxdy=xy</p><p></p><p></p><p>Differentiating on both sides w.r.t x we get,</p><p></p><p>Apply quotient rule</p><p></p><p></p><p>dx2d2y=x2x.dxdy−y.1</p><p></p><p>⇒x2x.dxdy−y</p><p></p><p>This can be written as</p><p></p><p>x1dxdy−x2y</p><p></p><p></p><p>But dxdy=xy</p><p></p><p></p><p>So substituting this we get,</p><p></p><p>x2y−x2y=0</p><p></p><p>Hence proved</p><p></p><p>$

Step-by-step explanation:

Given: x

m

y

n

=(x+y)

m+n

Take log on both side

logx

m

+logy

n

+(m+n)log=y(x+y)

mlogx+nlogy=(m+n)log(x+y)

Differentiating w.r.t x on both sides we get,

m.

x

1

+n

y

1

dx

dy

=(m+n).

(x+y)

1

(1+

dx

dy

)

x

m

+

y

n

dx

dy

=

x+y

m+n

(1+

dx

dy

)

dx

dy

(

y

n

−

x+y

m+n

)=

x+y

m+n

−

x

m

dx

dy

(

y(x+y)

n(x+y)−y(m+n)

)=

x(x+y)

x(m+n)−m(x+y)

On simplifying we get,

dx

dy

(

y

nx−my

)=

x

nx−my

∴

dx

dy

=

x

y

Differentiating on both sides w.r.t x we get,

Apply quotient rule

dx

2

d

2

y

=

x

2

x.

dx

dy

−y.1

⇒

x

2

x.

dx

dy

−y

This can be written as

x

1

dx

dy

−

x

2

y

But

dx

dy

=

x

y

So substituting this we get,

x

2

y

−

x

2

y

=0

Hence proved