log(base2) (x^2 - 1x - 52) =2
x(base1) = ?
x(base2) = ?
Solution to log equation ?
2016-04-16 1:51 am
回答 (3)
2016-07-27 5:50 pm
log 2 (x^2 -x -52) = 2
2 ^ log 2 (x^2 -x -52) = 2^2
(x^2 -x -52) = 4
x^2 -x -52 =4
x^2 -x - 56 = 0
(x - 8)(x + 7) = 0
x = 8 , x = -7
log 2 ((8)^2 - (8) -52) = 2
log 2 (64 -8 -52)
log 2 (4) = 2
2 =2 check
log 2 ((7)^2 - (7) -52) = 2
log 2 (49 - (-7) -52) =2
log 2 (49 + 7 -52) = 2
log 2 (56 - 52) = 2
log 2 (4) = 2
2 = 2 check
x = -7, 8
2 ^ log 2 (x^2 -x -52) = 2^2
(x^2 -x -52) = 4
x^2 -x -52 =4
x^2 -x - 56 = 0
(x - 8)(x + 7) = 0
x = 8 , x = -7
log 2 ((8)^2 - (8) -52) = 2
log 2 (64 -8 -52)
log 2 (4) = 2
2 =2 check
log 2 ((7)^2 - (7) -52) = 2
log 2 (49 - (-7) -52) =2
log 2 (49 + 7 -52) = 2
log 2 (56 - 52) = 2
log 2 (4) = 2
2 = 2 check
x = -7, 8
2016-04-16 2:02 am
log₂(x² - x - 52) = 2
Same start as the last time you asked this when I thought the base was 10. Instead of a base of 10, it's a base of 2.
x² - x - 52 = 2²
x² - x - 52 = 4
x² - x - 56 = 0
(x - 8)(x + 7) = 0
x = -7 and 8
Testing both to make sure that x² - x - 52 results in a positive number:
x² - x - 52
(-7)² - (-7) - 52 and 8² - 8 - 52
49 + 7 - 52 and 64 - 60
4 and 4
Both are positive, so both values of x are solutions:
x = -7 and 8
Same start as the last time you asked this when I thought the base was 10. Instead of a base of 10, it's a base of 2.
x² - x - 52 = 2²
x² - x - 52 = 4
x² - x - 56 = 0
(x - 8)(x + 7) = 0
x = -7 and 8
Testing both to make sure that x² - x - 52 results in a positive number:
x² - x - 52
(-7)² - (-7) - 52 and 8² - 8 - 52
49 + 7 - 52 and 64 - 60
4 and 4
Both are positive, so both values of x are solutions:
x = -7 and 8
2016-04-16 1:52 am
收錄日期: 2021-05-03 07:15:34
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