Determine the exact value for x if: log2 x + log2(x − 7) = log2 7x?
回答 (2)
2016-07-15 1:51 pm
✔ 最佳答案
log₂(x) + log₂(x - 7) = log₂(7x)log₂[x(x - 7)] = log₂(7x)
x(x - 7) = 7x
x² - 7x = 7x
x² - 14x = 0
x(x - 14) = 0
x = 0 (rejected) or x = 14
Hence, x = 14
2016-07-15 3:01 pm
log 2 (x) + log 2 (x+7) = log 2 7x
(x)(x+7) = 7x
x^2+7x = 7x
x^2 + 7x -7x = 0
x^2 = 0
x = 0, 0
The log of 0 base 2 does not exist nor does the log of -7 base 2
The answer is no solution.
(x)(x+7) = 7x
x^2+7x = 7x
x^2 + 7x -7x = 0
x^2 = 0
x = 0, 0
The log of 0 base 2 does not exist nor does the log of -7 base 2
The answer is no solution.
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