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
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
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