Logarithmic Equations

**Macleef** · December 12th 2007, 02:52 PM

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?

Answer: 7

**Plato** · December 12th 2007, 03:02 PM

$\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} $

**Macleef** · December 12th 2007, 03:13 PM

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?

**Plato** · December 12th 2007, 03:31 PM

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.
$\log _b \left( {\frac{1}{a}} \right) = \log _b \left( {a^{ - 1} } \right) = - \log _b \left( a \right)$ .

**Macleef** · December 12th 2007, 03:37 PM

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:

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

**Plato** · December 12th 2007, 03:43 PM

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.

**Macleef** · December 12th 2007, 04:27 PM

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

**earboth** · December 12th 2007, 09:00 PM

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.

Thread: Logarithmic Equations

LinkBack

Thread Tools

Display

Logarithmic Equations

Similar Math Help Forum Discussions

Logarithmic Equations

Logarithmic equations

Logarithmic equations

logarithmic equations

Logarithmic equations

Search Tags