Question

Guys please answer.......... ​

Answer 1

Given,

$\longrightarrow x^2+y^2=3xy$

Adding $2xy$ to both sides,

$\longrightarrow x^2+y^2+2xy=3xy+2xy$

$\longrightarrow (x+y)^2=5xy$

Taking logarithm on both sides,

$\longrightarrow \log\left[(x+y)^2\right]=\log(5xy)\quad\quad\dots(1)$

We have,

• $\log(a^b)=b\log a$

• $\log(ab)=\log a+\log b$

Thus (1) becomes,

$\longrightarrow\underline{\underline{2\log(x+y)=\log 5+\log x+\log y}}$

Logarithmic Identities,

$\longrightarrow\log(ab)=\log a+\log b$

$\longrightarrow\log\left(\dfrac{a}{b}\right)=\log a-\log b$

$\longrightarrow\log(a^b)=b\log a$

$\longrightarrow\log_ba=\dfrac{\log_ma}{\log_mb},\quad b\neq1$

$\longrightarrow a^b=m^{b\log_ma}$

Answer 2

Step-by-step explanation:

Given,

x² + y² = 3xy

Add 2xy on both sides

x² + y² + 2xy = 3xy + 2xy

(x + y)² = 5xy

Add log on both sides

log (x + y)² = log 5xy

2 log (x + y) = log 5 + log x + log y

Hence, proved.

Formulae applied here :

$\to$a² + b² + 2ab = (a + b)²

$\to$ log x$^m$ = m log x

$\to$ log xy = log x + log y