Question

If x3 + y3 =0 and x + y is not equal to 0 then prove that Log x+y= 1/2 (Logx + Logy+ log 3)

Answer 1

$\textbf{Given:}$

$\mathsf{x^3+y^3=0\;and\;x+y\;{\neq}\;0}$

$\textbf{To prove:}$

$\mathsf{log(x+y)=\dfrac{1}{2}(log\,x+log\,y+log\,3)}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{x^3+y^3=0}$

$\mathsf{(x+y)(x^2-xy+y^2)=0}$

$\mathsf{But\;x+y\;{\neq}\;0}$

$\mathsf{x^2-xy+y^2=0}$

$\mathsf{Adding\;bothsides\;by\;3xy}$

$\mathsf{(x^2-xy+y^2)+3xy=3xy}$

$\mathsf{x^2+2xy+y^2=3xy}$

$\mathsf{(x+y)^2=3xy}$

$\textsf{Taking logarithms on bothsides, we get}$

$\mathsf{log(x+y)^2=log(3xy)}$

$\textsf{Using prodiuct and power rule of logarithm,}$

$\mathsf{2\;log(x+y)=log\;3+log(xy)}$

$\implies\boxed{\mathsf{2\;log(x+y)=log\;3+log\;x+log\;y}}$

$\textbf{Find more:}$

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