# Math

Solve simultaneously log (x-1)+2logy=2log3 and logx+logy=log6

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1. log ( x -1 ) + 2 log y = 2 log 3

log ( x -1 ) = 2 log 3 - 2 log y

log ( x -1 ) = 2 ( log 3 - log y )

log ( x -1 ) = 2 log ( 3 / y )

log ( x -1 ) = log [ ( 3 / y )² ]

x - 1 = ( 3 / y )²

x - 1 = 9 / y²

x = 9 / y² + 1

log x + log y = log 6

log ( x ∙ y ) = log 6

x ∙ y = 6

x = 6 / y

Use fact:

x = x

9 / y² + 1 = 6 / y

Multiply both sides by y²

9 + y² = 6 y

Subtract 6 y to both sides

9 + y² - 6 y = 6 y - 6 y

y² - 6 y + 9 = 0

Factor the left hand side

( y - 3 )² = 0

y - 3 = √0

y - 3 = 0

y - 3 + 3 = 0 + 3

y = 3

x = 6 / y

x = 6 / 3

x = 2

Solution x = 2 , y = 3

Proof:

log ( x -1 ) + 2 log y = 2 log 3

log ( 2 -1 ) + 2 log 3 = 2 log 3

log 1 + 2 log 3 = 2 log 3

0 + 2 log 3 = 2 log 3

2 log 3 = 2 log 3

log x + log y = log 6

log 2 + log 3 = log 6

log ( 2 ∙ 3 ) = log 6

log 6 = log 6

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posted by Bosnian

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