Solve log(x^2 - 1) - 2 = log(x+1) for x.

[sfin]

log(x^2 - 1) - 2 = log(x+1)

log(x^2 - 1) - log(x+1) = 2

log x^2 - 1 / log(x+1) = 2

(x+1)(x-1) / (x+1) = 2

x-1 = 2

x = 3

My answer key shows x = 101. What am I missing?

 

[tkhunny]

Re: Solve for x.

log(x^2 - 1) - log(x+1) = 2

log x^2 - 1 / log(x+1) = 2

(x+1)(x-1) / (x+1) = 2

Both those steps are bad.

Stare at these until they soak in.

log(a) - log(b) = log(a/b) That is NOT log(a)/log(b)

log(a) = c ==> 10^c = a Where did that "10" come from?

 

[soroban]

Re: Solve for x.

Hello, sfin!

tkhunny is right . . . You started fine, then blew it.


\(\displaystyle \L\log(x^2\,-\,1)\,-\,2 \:=\:\log(x\,+\,1)\) . . Note that \(\displaystyle x\,>\,-1\)

\(\displaystyle \L\log(x^2\,-\,1)\,-\,\log(x\,+\,1)\:=\:2\)

\(\displaystyle \L\log\left(\frac{x^2\,-\,1}{x\,+\,1}\right)\:=\:2\)

Then: \(\displaystyle \L\:\frac{x^2\,-\,1}{x\,+\,1}\:=\:10^2\)

Factor: \(\displaystyle \L\:\frac{(x\,-\,1)(x\,+\,1)}{x\,+\,1}\:=\:100\)

Reduce: \(\displaystyle \L\:x\,-\,1\:=\:100\;\) **


Therefore: \(\displaystyle \L\:x\:=\:101\)

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**

We can cancel because \(\displaystyle x\,>\,-1\) . . . hence, \(\displaystyle x\,+\,1\:\neq\:0\)