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Solve a Logarithmic Equation

The question reads:

Solve the equation
ln(1 + x2) = 1 + 2 ln x,
giving your answer correct to 3 significant figures.

I managed to solve it for x and obtained ±0.763

But the mark scheme tells the examiner to deduct a mark for not implying that there is only ONE solution and that is +0.763.

Why is -0.763 not considered a solution?
Reply 1
Avatar for BabyMaths
BabyMaths
Original post by nonipaify
The question reads:

Solve the equation
ln(1 + x2) = 1 + 2 ln x,
giving your answer correct to 3 significant figures.

I managed to solve it for x and obtained ±0.763

But the mark scheme tells the examiner to deduct a mark for not implying that there is only ONE solution and that is +0.763.

Why is -0.763 not considered a solution?


What's ln(-0.763)?
Reply 2
Avatar for nerak99
nerak99
13
Original post by nonipaify
The question reads:

Solve the equation
ln(1 + x2) = 1 + 2 ln x,
giving your answer correct to 3 significant figures.

I managed to solve it for x and obtained ±0.763

But the mark scheme tells the examiner to deduct a mark for not implying that there is only ONE solution and that is +0.763.

Why is -0.763 not considered a solution?


The definition of a log is that it is a power to which the base is raised to equal a given number.

ln(-0.763) demands that there is a power to which e can be raised to equal -0.763.

You cannot raise e to any power at all and get a negative result and so you have to dismiss the negative value for x since there is a ln(x) in the original equation.

(I have had an email pointing out that including complex numbers as a power would allow a negative result but in the context of this post I thought best not to muddy the water with complex numbers).

Hope that this helps
(edited 9 years ago)
Reply 3
Avatar for nonipaify
nonipaify
OP
3
THANK you both for your quick replies. How stupid of me. Problem solved

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