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