Solve. log6x+log6(x−1)=1 ?
回答 (5)
2019-03-05 9:53 pm
log₆(x) + log₆(x - 1) = 1
log₆[x(x -1)] = log₆(6)
x(x - 1) = 6
x² - x - 6 = 0
(x - 2)(x + 3) = 0
x = 2 or x = -3 (rejected)
Hence, x = 2
log₆[x(x -1)] = log₆(6)
x(x - 1) = 6
x² - x - 6 = 0
(x - 2)(x + 3) = 0
x = 2 or x = -3 (rejected)
Hence, x = 2
2019-03-05 9:59 pm
log (base 6) x + log (base 6) (x - 1) = 1
log (base 6) (x * (x - 1)) = 1
6^1 = x * (x - 1)
x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = -2, 3
The logarithm function is defined for (0, +∞) => x cannot be -2
x = 3
log (base 6) (x * (x - 1)) = 1
6^1 = x * (x - 1)
x^2 - x - 6 = 0
(x - 3)(x + 2) = 0
x = -2, 3
The logarithm function is defined for (0, +∞) => x cannot be -2
x = 3
2019-03-05 9:54 pm
is this -- log(base6)x + log(base6)(x-1) = 1 ?
x(x-1) = 6
x^2 - x - 6 = 0
(x-3)(x+2) = 0 .... x = 3 and x = -2
cannot = -2, log(-2) is undefinded for reals
x = 3 <<< answer
x(x-1) = 6
x^2 - x - 6 = 0
(x-3)(x+2) = 0 .... x = 3 and x = -2
cannot = -2, log(-2) is undefinded for reals
x = 3 <<< answer
2019-03-05 9:56 pm
log a + log b = log a*b
x(x-1) = 6^1
x² - x - 6 = 0
x = [1 ± √(1² + 24)]/2
x = 3 accepted answer
x(x-1) = 6^1
x² - x - 6 = 0
x = [1 ± √(1² + 24)]/2
x = 3 accepted answer
2019-03-05 11:58 pm
--
Let log be log to base 6
log [ x / ( x - 1 ) ] = 1
x / ( x - 1 ) = 6
x = 6x - 6
6 = 5x
x = 6/5
Let log be log to base 6
log [ x / ( x - 1 ) ] = 1
x / ( x - 1 ) = 6
x = 6x - 6
6 = 5x
x = 6/5
收錄日期: 2021-05-01 22:29:23
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