J's answer to Shams's Secondary 4 A Maths Singapore question.
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log y + log x = log (y + x)
log yx = log (y + x)
Since the log bases are the same, the arguments are equal.
yx = y + x
yx - y = x
y(x - 1) = x
y = x / (x - 1)
y = (x - 1 + 1) / (x - 1)
y = 1 + 1/(x - 1)
log yx = log (y + x)
Since the log bases are the same, the arguments are equal.
yx = y + x
yx - y = x
y(x - 1) = x
y = x / (x - 1)
y = (x - 1 + 1) / (x - 1)
y = 1 + 1/(x - 1)
Date Posted:
3 years ago
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