log-x-y-1-log-2-x-log-4-y-2-4-

Question Number 42289 by mondodotto@gmail.com last updated on 22/Aug/18

\log (x + y) = 1

\log_{2} x + \log_{4} y^{2} = 4

Commented by math khazana by abdo last updated on 22/Aug/18

\Rightarrow x + y = e a n d \frac{l n (x)}{l n (2)} + \frac{l n (y^{2})}{l n (4)} = 4 \Rightarrow

x + y = e a n d l n (x) + l n (y) = 4 l n (2) = l n (2^{4}) \Rightarrow

x + y = e a n d l n (x y) = l n (16) \Rightarrow

x + y = e a n d x y = 16 \Rightarrow x a n d y a r e s o l u t i o n o f t h e

e q u a t i o n u^{2} - e u + 16 = 0

Δ = e^{2} - 4 (16) = e^{2} - 64

i f x a n d y a r e r e a l \Rightarrow n o s o l u t i o n i f x a n d y a r e

c o m p l e x \Rightarrow u_{1} = \frac{e + i \sqrt{64 - e^{2}}}{2}

u_{2} = \frac{e - i \sqrt{64 - e^{2}}}{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Aug/18

x + y = e^{1} = e

\frac{l n x}{l n 2} + \frac{2 l n y}{2 l n 2} = 4

l n x + l n y = 4 l n 2

l n (x y) = l n 2^{4}

x y = 16

x + y = e

x (e - x) = 16

x e - x^{2} - 16 = 0

x^{2} - x e + 16 = 0

x = \frac{e \pm \sqrt{e^{2} - 64}}{2} = \frac{e \pm i \sqrt{64 - e^{2}}}{2}

y = e - \frac{e \pm \sqrt{e^{2} - 64}}{2} = \frac{e \underset{+}{-} \sqrt{e^{2} - 64}}{2} = \frac{e \underset{+}{-} i \sqrt{64 - e^{2}}}{2}

o r

i f l o g (x + y) = 1 {l o g}_{10} (x + y) = 1

x + y = 10

t h e n x = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm 6}{2} = 8 a n d 2

s o y = 2 a n d 8