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$\ln(x + 1) = 2 + \ln(x - 1)$; solve for $x$.

From there I get $\ln \frac{x+1}{x-1} = 2.$

Am I headed in the right direction, in our examples we would exponentiate both sides, does that still stand for this even though there's only a 2 on the right hand side?

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    Minor comment: The subtraction is fine of course, but myself I would prefer to take the exponential immediately, get $x+1=(e^2)(x-1)$.2011-12-06

3 Answers 3

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You are right. From

$\ln \frac{x+1}{x-1} = 2.$

we can write it as

$e^{\ln \frac{x+1}{x-1}}=e^{2} $

then

$\frac{x+1}{x-1}=e^{2}$

You can find the solution for $x$

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Yes, you use the fact that $\ln u=v\iff u=e^v.$ So applying this to what you have (with $u={x+1\over x-1}$ and $v=2$): $ \tag{1}{x+1\over x-1}=e^2. $ Multiplying both sides by $x-1$ gives $ \tag{2}x+1=e^2(x-1). $ (note, here that $x=1$ is not a solution of (2); so (1) and (2) are equivalent equations)

Solving for $x$ in (2):

$\eqalign{ &x+1= e^2 x-e^2\cr &\iff x-e^2x =-1-e^2\cr &\iff x(1-e^2)=-1-e^2\cr &\iff x= {-1-e^2\over 1-e^2} } $ Or $ x={e^2+1\over e^2-1}. $

When solving logarithmic equations, you sould always check your answers. In particular, check that you don't wind up taking the logarithm of a non-positive quantity.

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Why not take the exponential of both sides immediately? We get $x+1=(e^2)(x-1),$ and it's over.

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    That would be correct David, as we just went over the rules o$f$ logs the previous week.2011-12-06