# Logs :(

• Dec 10th 2012, 07:54 AM
Twinkie
Logs :(
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 AM
topsquark
Re: Logs :(
Quote:

Originally Posted by Twinkie
1. write as a single logarithm in simplest form
(Log 16x^8)/4 - (Log 27x)/3

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

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

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

Skipping a couple of steps, you should arrive at
$2 \left [ \frac{ln(x)}{ln(16)} \right ] - \frac{1}{3} \left [ \frac{ln(x)}{ln(27)} \right ]$

Which has the final form, as simply as I can write it:
$\left [ \frac{1}{2~ln(2)} - \frac{1}{9~ln(3)} \right ] ln(x)$

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

-Dan
• Dec 10th 2012, 08:57 AM
topsquark
Re: Logs :(
Quote:

Originally Posted by Twinkie
2. isolate x using laws of logarithims
3 Logx(4)=2

$\log_x(4) = 2$

$x^{\log_x(4)} = x^2$

$4 = x^2$

And you can finish from there. Watch out for the extra (wrong) solution that comes from this.

-Dan
• Dec 10th 2012, 09:14 AM
Twinkie
Re: Logs :(
Ok thankyou for your help. :)
• Dec 10th 2012, 01:44 PM
bjhopper
Re: 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 PM
Twinkie
Re: 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 PM
topsquark
Re: Logs :(
Quote:

Originally Posted by bjhopper
1. log 16x^8/4 - log27x/3
2.log4x^8 - log 9x

The division of 4 in the first term is outside the logarithm. So you can't divide through.

-Dan
• Dec 10th 2012, 05:36 PM
topsquark
Re: Logs :(
Quote:

Originally Posted by Twinkie
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

If the bases on all the logs are the same then we can do this rather easily. Let's get everything to one side:
$2\log_6(x) - \log_6(9) - \log_6(24) = 0$

Let's rewrite that first term:
$\log_6(x^2) - \log_6(9) - \log_6(24) = 0$

Now we can use
$\log_a(p) + \log_a(q) = \log_a(pq)$

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

(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 PM
bjhopper
Re: Logs :(
Quote:

Originally Posted by Twinkie
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

Twinkie,
1.log base 6 (x)^3= log b6 (9*24) =log b6 (216)
2. x^3 = 216
3. x=6
• Dec 10th 2012, 07:22 PM
bjhopper
Re: Logs :(
Quote:

Originally Posted by topsquark
The division of 4 in the first term is outside the logarithm. So you can't divide through.

-Dan

Dan, That was a mistake but when I redo the question I come up with log 2/3 X^5/3 as the simplified log
• Dec 11th 2012, 08:34 AM
Twinkie
Re: Logs :(
Ok thanks! :)