Question

(log x² – log x) . log 1/x + (log x)² = 0​

Answer 1

Step-by-step explanation:

(2 log x - log x) *(log 1 - log x) +2 logx

log x *(- logx) + 2log x (log1=0)

(-log x) ^2+ 2log x

-2log x + 2 log x =0

Answer 2

Answer:

(log x² – log x) . log 1/x + (log x)² = 0

Step-by-step explanation:

(log x² – log x) . log 1/x + (log x)²

$=( 2 log(x) - log(x) ).( log(1) - log(x) ) + {( log(x) )}^{2}$

$= ( log(x) ).(0 - log(x) ) + {( log(x)) }^{2}$

$= log(x) .( - log(x) ) + {( log(x)) }^{2}$

$= - {( log(x) )}^{2} + {( log(x) )}^{2}$

$= 0$

Hence proved.

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