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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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Yes, you get $e^2$ on the right... –  J. M. Dec 6 '11 at 16:12
Yes, you are doing well. After a short while you will get a linear equation for $x$. –  André Nicolas Dec 6 '11 at 16:12
Yes, you are in the right direction. Now take exponentials on both sides to eliminate the $\ln$. –  Arturo Magidin Dec 6 '11 at 16:13
Minor comment: The subtraction is fine of course, but myself I would prefer to take the exponential immediately, get $x+1=(e^2)(x-1)$. –  André Nicolas Dec 6 '11 at 16:23
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3 Answers

up vote 4 down vote accepted

You are right. From

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

we can write it as

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



You can find the solution for $x$

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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 a better way to do it. Most College Algebra texts I've seen have you "collect the logs" first. I suppose the authors want to give the student practice using the rules of logarithms. –  David Mitra Dec 6 '11 at 16:39
That would be correct David, as we just went over the rules of logs the previous week. –  erimar77 Dec 6 '11 at 16:44
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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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