Logarithmic Equations

**castle** · July 15th 2011, 02:21 PM

I've tried this one and feel my answer is correct but my textbook says otherwise.

$log_8 (x^2 - 1) - log_8(x - 1) = 2$

I get an answer of 63, but my book says 15. What am I doing wrong?

**Plato** · July 15th 2011, 02:31 PM

Originally Posted by castle

I've tried this one and feel my answer is correct but my textbook says otherwise.
$log_8 (x^2 - 1) - log_8(x - 1) = 2$
I get an answer of 63, but my book says 15. What am I doing wrong?

You are correct, the book is not.

**mr fantastic** · July 15th 2011, 02:35 PM

Originally Posted by castle

I've tried this one and feel my answer is correct but my textbook says otherwise.

$log_8 (x^2 - 1) - log_8(x - 1) = 2$

I get an answer of 63, but my book says 15. What am I doing wrong?

Your mistake is to think that you're wrong and the book is right.

It is simple to check each answer and see which one is correct ....

**Prove It** · July 15th 2011, 09:02 PM

First note that $\displaystyle x \neq 1$ because $\displaystyle \log{0}$ is undefined.

$\displaystyle \begin{align*} \log_8{(x^2-1)} - \log_8{(x-1)} &= 2 \\ \log_8{\left(\frac{x^2 - 1}{x - 1}\right)} &= 2 \\ \frac{x^2 - 1}{x - 1} &= 8^2 \\ \frac{x^2 - 1}{x - 1} &= 64 \\ x^2 - 1 &= 64(x - 1) \\ x^2 - 1 &= 64x - 64 \\ x^2 - 64x + 63 &= 0 \\ (x - 63)(x - 1) &= 0 \\ x - 63 &= 0 \\ x &= 63 \end{align*}$

You are correct.

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