Ally Chua's answer to LockB's Secondary 4 A Maths Singapore question.
when a^2 = b,
have to consider both positive & negative roots,
so a =±sqrt[b].
(lnx)^2 = ln20
ln x = ±sqrt[ln20]
x = e^(±sqrt[ln20])
= 5.6453 or 0.1771
both values can satisfy the original equation.
there are 2 possible solutions!
have to consider both positive & negative roots,
so a =±sqrt[b].
(lnx)^2 = ln20
ln x = ±sqrt[ln20]
x = e^(±sqrt[ln20])
= 5.6453 or 0.1771
both values can satisfy the original equation.
there are 2 possible solutions!
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