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$\log_7 (x^2-1) - \log_7 (x-1) = 2$

$\log_7 49 = 2$

=> $\log_7 (x^2-1) - \log_7 (x-1) = \log_7 49$

=> $\frac{(x^2-1)}{x-1} = 49$

=> $x^2 -1 = 49(x -1)$

=> $x^2 -1 = 49x -49$

=> $x^2 - 49x + 48 = 0$

The answer in the book is 48 so I'm obviously doing this wrong.

3SAT
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dagda1
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    You will have two roots of that quadratic; 48 and 1. However at various steps in your workings, the expressions are undefined for the value 1, so you are left with $x = 48$. – Edward Evans Mar 23 '16 at 08:29

2 Answers2

1

Since $x=1$ is excluded, your equation $$ \frac{(x^2-1)}{x-1} = 49 $$ is just $$ x+1=49. $$

Olivier Oloa
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1

We have that $x>1$ ($\log_7x$ is only defined if $x>0$) and $$ \log_7(x^2-1)-\log_7(x-1)=\log_7\biggl(\frac{x^2-1}{x-1}\biggr)=\log_7\biggl(\frac{(x-1)(x+1)}{x-1}\biggr)=\log_7(x+1)=2. $$ Hence, $x+1=49$ and $x=48$.

Cm7F7Bb
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