Question

log (x+y)/5=1/2(logx+logy) then what is x/y+y/x​

Answer 1

EXPLANATION.

⇒ ㏒(x + y)/5 = 1/2(㏒ x + ㏒ y).

As we know that,

Formula of :

⇒ ㏒ₐN^(α) = α ㏒ₐN (α any real numbers).

Using this formula in the equation, we get.

We can write equation as,

⇒ ㏒(x + y)/5 = (㏒ x + ㏒ y)^(1/2).

⇒ ㏒(x + y)/5 = √(㏒ x + ㏒ y).

⇒ ㏒(x + y)/5 = ㏒ √xy.

⇒ (x + y)/5 = √xy.

⇒ (x + y) = 5√xy.

Squaring on both sides of the equation, we get.

⇒ (x + y)² = (5√xy)².

⇒ x² + y² + 2xy = 25xy.

⇒ x² + y² = 25xy - 2xy.

⇒ x² + y² = 23xy.

⇒ x² + y²/xy = 23.

⇒ (x²/xy) + (y²/xy) = 23.

⇒ (x/y) + (y/x) = 23.

                                                                                                                 

MORE INFORMATION.

Properties of logarithms.

Let M and N arbitrary positive numbers such that a > 0, a ≠ 1, b > 0, b ≠ 1 then,

(1) = ㏒ₐMN = ㏒ₐM + ㏒ₐN.

(2) = ㏒ₐ(M/N) = ㏒ₐM - ㏒ₐN.

(3) = ㏒ₐN^(α) = α ㏒ₐN (α any real numbers).

(4) = ㏒ₐ^(β)N^(α) = (α/β)㏒ₐN (α ≠ 0, β ≠ 0).

(5) = ㏒ₐN = ㏒_{b}N/㏒_{b}a.

(6) = ㏒_{b}a . ㏒ₐ b = 1 ⇒ ㏒_{b}a = 1/㏒ₐb.

(7) = e^(㏑a)ˣ = aˣ.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:log\bigg[\dfrac{x + y}{5} \bigg] = \dfrac{1}{2}\bigg[logx \: + \: logy\bigg]$

can be rewritten as

$\rm :\longmapsto\:2log\bigg[\dfrac{x + y}{5} \bigg] = \bigg[logx \: + \: logy\bigg]$

We know,

$\boxed{ \bf{ \:bloga = log {a}^{b}}}$

and

$\boxed{ \bf{ \:logx + logy = log(xy)}}$

So, using these Identities, we get

$\rm :\longmapsto\:log\bigg[\dfrac{x + y}{5} \bigg]^{2} = \bigg[logxy \bigg]$

So, on comparing, we get

$\rm :\longmapsto\:\bigg[\dfrac{x + y}{5} \bigg]^{2} = xy$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + {y}^{2} + 2xy}{25} = xy$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + 2xy = 25xy$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 25xy - 2xy$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} = 23xy$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + {y}^{2} }{xy} = 23$

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{xy} + \dfrac{ {y}^{2} }{xy} = 23$

$\bf\implies \:\dfrac{x}{y} + \dfrac{y}{x} = 23$

Additional Information :-

$\boxed{ \bf{ \: log_{x}(x) = 1}}$

$\boxed{ \bf{ \: log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y} }}$

$\boxed{ \bf{ \: {x}^{ log_{x}(y) } = y}}$

$\boxed{ \bf{ \: {x}^{z log_{x}(y) } = {y}^{z} }}$

$\boxed{ \bf{ \: {e}^{logx} = x}}$

$\boxed{ \bf{ \: {e}^{ylogx} = {x}^{y} }}$