Math, asked by balasagar0786, 9 months ago

If log (x+y/5)=1/2 (logx+logy) find the value of x/y+y/x​

Answers

Answered by vmendadalavenkatasat
11

Answer:

23

Step-by-step explanation:log (x+y /5)=1/2 (logx+logy)

log (x+y/5)^2=(logx+logy) (a+b)^2=a^2+b^2+2ab

log (x^2+y^2+2xy/25)=(logx+logy)

log (x^2+y^2+2xy)=log25xy

Remove log on both sides,

x^2+y^2+2xy=25xy

x^2+y^2=25xy-2xy

x^2+y^2=23xy

Dividing with 2xy, on both sides

x^2+y^2/xy=23xy/xy

x/y+y/x=23

Answered by amansharma264
18

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) = ㏒M - ㏒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.

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