Question

prove that (log x^2 - log x). log(1/x) +(log x)^2 =0​

Answer 1

Answer:

I am getting the answer but I had tried half

Step-by-step explanation:

given : prove that (log x^2 - log x). log(1/x) +(log x)^2 =0

(log x^2 - log x). log(1/x) +(log x)^2 =0

( log x² - log x ) ( log 1/x ) + log x²=0

(2logx-logx ) ( log 1/x ) + log x²=0

(log x ) ( log 1-logx) +2logx = 0

( log 1-logx) = -2logx/logx

( log 1-logx) = -2

I am getting the answer but I had tried half

Answer 2

$\sf{\log _{10}\left(x+\sqrt{x^2-1}\right)+\log _{10}\left(x-\sqrt{x^2-1}\right)}$

Apply Log Rule : $\bf{\log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)}$

$\sf{=\log _{10}\left(\left(x+\sqrt{x^2-1}\right)\left(x-\sqrt{x^2-1}\right)\right)}$

Apply Difference of Two Squares Formula : $\bf{(a+b)(a-b)=a^2-b^2}$

$\sf{=\log _{10}\left(x^2-\left(\sqrt{x^2-1}\right)^2\right)}$

$\sf{=\log _{10}\left(x^2-\left(x^2-1\right)\right)}$

$\sf{=\log _{10}\left(x^2-x^2+1\right)}$

$\sf{=\log _{10}\left(1\right)}$

Apply Log Rule : $\bf{\log _a\left(1\right)=0}$

$\sf{=0}$

Hence Proved !