$log (x^{2}) + log (y^{2})$ is equal to

log (x^{2} + y^{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 log (x y)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 log (x + y)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

log \frac{x^{2}}{y^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 log (x y)$
$log (x^{2}) + log (y^{2})$
$= log (x^{2} \times y^{2}) = log (x y)^{2} = 2 log x y$

Suggest Corrections

Similar questions

logx+ log( xy⁸)/(log x)²+(log y)²=2 & logy+log(x⁸/y)/(log x)²+(log y)²=0( where base of log is 10 ) compute the product of(xy).

\frac{(l o g x - l o g y) (l o g x^{2} + l o g y^{2})}{(l o g x^{2} - l o g y^{2}) (l o g x + l o g y)}

l o g_{2} x + l o g_{x} 2 = \frac{10}{3} = l o g_{2} y + l o g_{y} 2 a n d x \neq y, t h e n x + y^{3}

{log}_{2} x + {log}_{x} 2 = \frac{10}{3} = {log}_{2} y + {log}_{y} 2

x = y

x + y

\int \frac{\sqrt{x^{2} + 1} [log (x^{2} + 1) - 2 log x]}{x^{4}} d x

Join BYJU'S Learning Program