1. ## Logarithmic Equations

Solve:
$log_8(x-3) - log_8 (3x + 11) = log_8 (\frac{1}{x+1})$

I have this:

$log_8(\frac {x-3}{3x + 11}) = log_8 (\frac{1}{x+1})$

$log_8(-2x -14) = log_8 (-x)$

But I'm not sure my second step is right...how would you solve this equation?

2. $\begin{array}{l}
\log _8 \left( {x - 3} \right) - \log _8 \left( {3x + 11} \right) = \log _8 \left( {\frac{1}{{x + 1}}} \right) = - \log _8 \left( {x + 1} \right) \\
\log _8 \left( {\frac{{\left( {x - 3} \right)\left( {x + 1} \right)}}{{3x + 11}}} \right) = 0 \\
\end{array}
$

3. Originally Posted by Plato
$\begin{array}{l}
\log _8 \left( {x - 3} \right) - \log _8 \left( {3x + 1} \right) = \log _8 \left( {\frac{1}{{x + 1}}} \right) = - \log _8 \left( {x + 1} \right) \\
\log _8 \left( {\frac{{\left( {x - 3} \right)\left( {x + 1} \right)}}{{3x + 1}}} \right) = 0 \\
\end{array}
$
How did you get a $-log$, if you were to move the numerator of 1, wouldn't it be just $log$?
Also, what happened to the 11 in the denominator of the first log?

4. Originally Posted by Macleef
what happened to the 11 in the denominator of the first log?
That was a mere typo! I have corrected it.
In any event it would not change the basic idea. You really should know the basic properties.

Originally Posted by Macleef
How did you get a $-log$, if you were to move the numerator of 1, wouldn't it be just $log$?
Again you should know basic properties.
$\log _b \left( {\frac{1}{a}} \right) = \log _b \left( {a^{ - 1} } \right) = - \log _b \left( a \right)$.

5. Could you show me how to solve this equation step by step?
I don't see how $log_8 (\frac {1}{x+1}) = -log_8 (x+1)$
As for the negative exponent law, did you take the exponent of -1 from (x+1), is that what you did?

And how would you solve for x in:

$(\frac {x^{2} - 2x - 3}{3x + 11}) = 0$

6. Originally Posted by Macleef
Could you show me how to solve this equation step by step?
I don't see how $log_8 (\frac {1}{x+1}) = -log_8 (x+1)$
As for the negative exponent law, did you take the exponent of -1 from (x+1), is that what you did?
I think that you need more help than you can expect from here.
Go talk to your instructor. You need live one-on-one help.

7. No, I don't need extra help.
I got the question, I forgot to cross-multiply in my way.

8. Originally Posted by Macleef
Solve:
$log_8(x-3) - log_8 (3x + 11) = log_8 (\frac{1}{x+1})$

...
Hello,

1. I agree with Plato that you only have very basic knowledge about powers and logarithms. Therefore you have to learn by heart until tomorrow the content of the attachment.

2.
$log_8(x-3) - log_8 (3x + 11) = log_8 (\frac{1}{x+1})~\iff~$ $\log_8\left(\frac{x-3}{3x+11}\right) = \log_8(1) - \log_8(x+1)~,~x > 3$

Now de-logarithmize(?):

$\frac{x-3}{3x+11} = \frac{1}{x+1}~\iff~(x-3)(x+1)=3x+11 ~\iff~ x^2-5x-14=0$

Solve this quadratic equation for x. I've got: $x = -2~\vee~x = 7$

x = -2 doesn't belong to the domain of the equation thus you have the solution x = 7.

So you are right - but I really would like to know how you got this solution.