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secondary 4 | A Maths
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Date Posted: 3 years ago
Views: 182
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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)
Thx
Thx
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