Question

how do you fined the simplified version of 1/2 log3 x-2 log3 y-4 log3 z

Answer 1

When you have logs with the same base added to each other, you can combine them into a single log of the same base with their arguments multiplied (just remember to put the coefficients as exponents first). When they're subtracted, you can divide their arguments.

Answer 2

This?:\[\frac{1}{2}\log_3{(x-2)}-2\log_3{(y-4)}-4\log_3{(z)}\]

Answer 3

I think s/he meant:
\[\tfrac{1}{2}\log_3 x - 2\log_3 y -4\log_3 z\]

Answer 4

ya thats what she meant, but i dont understand how you would simplify it

Answer 5

Oh, yeah. So the first step @mboorstin described looks like going from \[\frac{1}{2}\large\log_3{x}-2\log_3{y}-4\log_3{z}=\large\log_3{x^{\frac{1}{2}}}-\log_3{y^2}-\log_3{z^4}\]

Answer 6

Then you use the fact that: \[\log a - \log b=\log\frac{a}{b}\]to manipulate\[\large\log_3{x^{\frac{1}{2}}}-\log_3{y^2}-\log_3{z^4}=\log_3\frac{x^{\frac{1}{2}}}{y^2z^4}\]

Answer 7

Now use the fact that log a + log b = log ab

Answer 8

ok so there are formulas for this?

Answer 9

Yes. There are a few basic log formulas you should know (if you think about how exponents work, they're pretty easy to derive):
\[\log a + \log b = \log ab\\
\log a - \log b = \log \tfrac{a}{b}\\
a \log b = \log \left(b^a\right)\\
\log_a b = \frac{\log_k b}{\log_k a}\\
a^{\log_a b} = b
\]

Answer 10

well thank you so much :) i have somewhat conquered the ways of log lol