Ok, I'm having some issues with these questions. So could you show me step by step on how to come to an answer.

1. write as a single logarithm in simplest form

(Log 16x^8)/4 - (Log 27x)/3

2. isolate x using laws of logarithims

3 Logx(4)=2

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- Dec 10th 2012, 07:54 AMTwinkieLogs :(
Ok, I'm having some issues with these questions. So could you show me step by step on how to come to an answer.

1. write as a single logarithm in simplest form

(Log 16x^8)/4 - (Log 27x)/3

2. isolate x using laws of logarithims

3 Logx(4)=2 - Dec 10th 2012, 08:54 AMtopsquarkRe: Logs :(
I'm afraid that I don't understand the final goal: "simplest form." So I'll show you how I would do the problem and you can make your own decision if the answer is simple enough.

We're going to be using the change of base formula:

$\displaystyle \log_a(x) = \frac{\log_b(x)}{\log_b(a)}$

I'm going to choose my favorite base, e.

So

$\displaystyle \frac{1}{4}\log_{16}(x^8) - \frac{1}{3} \log_{27}(x)$

Skipping a couple of steps, you should arrive at

$\displaystyle 2 \left [ \frac{ln(x)}{ln(16)} \right ] - \frac{1}{3} \left [ \frac{ln(x)}{ln(27)} \right ]$

Which has the final form, as simply as I can write it:

$\displaystyle \left [ \frac{1}{2~ln(2)} - \frac{1}{9~ln(3)} \right ] ln(x)$

The only other thing I can think of would be to subtract the fractions.

-Dan - Dec 10th 2012, 08:57 AMtopsquarkRe: Logs :(
- Dec 10th 2012, 09:14 AMTwinkieRe: Logs :(
Ok thankyou for your help. :)

- Dec 10th 2012, 01:44 PMbjhopperRe: Logs :(
1. log 16x^8/4 - log27x/3

2.log4x^8 - log 9x

3.log4x^8/9x

4.log 4/9x^7 - Dec 10th 2012, 05:15 PMTwinkieRe: Logs :(
That kind of question makes sence now, but I have anouther question that I'm not sure how to do.

Solve.

3log(6) X= log(6) 9+ log(6) 24 - Dec 10th 2012, 05:31 PMtopsquarkRe: Logs :(
- Dec 10th 2012, 05:36 PMtopsquarkRe: Logs :(
If the bases on all the logs are the same then we can do this rather easily. Let's get everything to one side:

$\displaystyle 2\log_6(x) - \log_6(9) - \log_6(24) = 0$

Let's rewrite that first term:

$\displaystyle \log_6(x^2) - \log_6(9) - \log_6(24) = 0$

Now we can use

$\displaystyle \log_a(p) + \log_a(q) = \log_a(pq)$

and

$\displaystyle \log_a(p) - \log_a(q) = \log_a \left ( \frac{p}{q} \right )$

(Note: These are only true when the base a is the same on all log functions.)

Can you finish from here?

-Dan - Dec 10th 2012, 06:36 PMbjhopperRe: Logs :(
- Dec 10th 2012, 07:22 PMbjhopperRe: Logs :(
- Dec 11th 2012, 08:34 AMTwinkieRe: Logs :(
Ok thanks! :)