Could I please have some help to solve this logarithm.
log₅(2x+1) + log₅(x-1) = 1
log₅(2x+1) + log₅(x-1) = 1
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Let log be log to base 5
log [ ( 2x + 1 ) ( x - 1 ) ] = 1
2 x ² - x - 1 = 5
2 x ² - x - 6 = 0
( 2x + 3 ) ( x - 2 ) = 0
x = 2 is acceptable
I like your subscript 5.
What annoys me is that I don`t know how to get it ! ! !
log [ ( 2x + 1 ) ( x - 1 ) ] = 1
2 x ² - x - 1 = 5
2 x ² - x - 6 = 0
( 2x + 3 ) ( x - 2 ) = 0
x = 2 is acceptable
I like your subscript 5.
What annoys me is that I don`t know how to get it ! ! !