Logarithms Properties/Identities/Etc.

**JSB1917** · September 18th 2009, 03:38 PM

Need some help on these, just to make sure I understand what I'm doing...

1) If $\log_ba = x$ then $\log_ab =$

$\frac{1}{\log_ab}$

2) If $\log_ba = x$ then $\log_\frac{1}{b}a =$

$-\log_ba$

3) $\log(2 - x) + \log(3 - x) = \log12$ solve for x

$x = -1$

4) $y = \log_bx + \log_b(x - 6)$ solve for x

$b^y = x^2 - 6x$

then...

would the quadratic formula be used?

$x^2 - 6x - b^y = 0$

??

**pickslides** · September 18th 2009, 03:48 PM

Originally Posted by JSB1917

4) $y = \log_bx + \log_b(x - 6)$ solve for x

$b^y = x^2 - 6x$

then...

would the quadratic formula be used?

$x^2 - 6x - b^y = 0$

??

To solve for x yes, use $a=1, b=-6$ and $c = - b^y$

**JSB1917** · September 18th 2009, 04:01 PM

Originally Posted by pickslides

To solve for x yes, use $a=1, b=-6$ and $c = - b^y$

Ok, so it would come out to...

$x = \frac {6 \pm \sqrt{36 + 4b^y}}{2}$

$x > 6$

**JSB1917** · September 22nd 2009, 07:23 AM

Originally Posted by JSB1917

Need some help on these, just to make sure I understand what I'm doing...

1) If $\log_ba = x$ then $\log_ab =$

$\frac{1}{\log_ab}$

2) If $\log_ba = x$ then $\log_\frac{1}{b}a =$

$-\log_ba$

Ok, I'm confused. My teacher marked both of these wrong. Instead he wanted $\frac {1}{x}$ for the first and -x for the second, but isn't what I put essentially the logical equivalence of the answers?

I got a zero for both of these. I had to turn back in the HW sheet because he completely missed a problem and gave a zero for it and I didn't have time to think about these two at the moment.

I can't understand why I shouldn't get at least 1 point for both of the problems, even though he wanted the work shown.

**mathaddict** · September 22nd 2009, 07:37 AM

Originally Posted by JSB1917

Need some help on these, just to make sure I understand what I'm doing...

1) If $\log_ba = x$ then $\log_ab =$

2) If $\log_ba = x$ then $\log_\frac{1}{b}a =$

3) $\log(2 - x) + \log(3 - x) = \log12$ solve for x

(1) $\frac{1}{\log_ab}=x \Rightarrow \log_ab=\frac{1}{x}$

(2) $\frac{1}{log_a\frac{1}{b}}=\frac{1}{-log_ab}=-\frac{1}{x}$

(3)log (2-x)(3-x)=log 12

x^2-5x-6=0

(x-6)(x+1)=0

Continue solving here .

**JSB1917** · September 22nd 2009, 08:03 AM

Originally Posted by mathaddict

..

wait what?

Also I noticed a typo earlier, it's supposed to be for number one.

$\frac {1}{\log_b a}$

**mathaddict** · September 22nd 2009, 08:07 AM

Originally Posted by JSB1917

wait what?

Also I noticed a typo earlier, it's supposed to be for number one.

$\frac {1}{\log_b a}$

edited , so clear now ?

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