Question

simplified version of 1/2 logsub 3 x + 2 log sub 3 y + 4 logsub 3 z

Answer 1

Did you mean this?

\[(1/2)*\log_3(x) + (2)*\log_3(y) + (4)*\log_3(z) \]

Answer 2

There are 3 different ways to simplify that I can think of, would you like the simplest denominator?

Answer 3

no. i mean 1/2 log sub 3 x+2 log sub3 y+4 log sub3 z and sure

Answer 4

?

Answer 5

In base ten (decimal # system), it simplifies to:
\[( \log(x) + 4\log(y) + 8\log(z) ) / \log(9)\]

A helpful rule for logarithms that is relevant here:
\[\log_A X = (\log X) / (logA)\]

For example:
\[\log_32 = \log 2 / \log 3\]

The default if there's no base specified is log base 10.

Does that help?

Answer 6

the answer should have log base 3 = the sqrt of x something

Answer 7

?

Answer 8

this is the problem ((1)/(2))log[3]x+2log[3]y+4log[3]z

Answer 9

hereq

Answer 10

do u know that a * log (x) = log (x^a)

Answer 11

?

Answer 12

what?

Answer 13

example
5 * log 3 = log (3^5)

Answer 14

Is this listed an equation you're solving for? I was under the impression it was an expression you needed simplified... Based on what you initially wrote as your question :-/

Answer 15

ok my answer choices are

Answer 16

If it's like this:
\[\log_3 x = ((1)/(2))\log_3 x+2 \log_3 y +4 \log_3 z\]

This can be solved in terms of log base 3

Answer 17

log base 3 sqrt x/ y^2z^4
log base 3 sqrt x y^2/ z^4
log base 3 sqrt x z^4/y^2
log base 3 sqrt x y^2 z^4

Answer 18

4th one

Answer 19

why? @agentx5 is this correct

Answer 20

This will be kind of messy, but these following equations are identical.

For base 3:
\[(y^2 z^4 \log(\sqrt(x)))/(\log(3)) = (\log(x))/(2 \log(3))+(2 \log(y))/(\log(3))+(4 \log(z))/(\log(3))\]

For base 10:
\[y^2 z^4 \log(x) = \log(x)+4 \log(y)+8 \log(z)\]

Read what I wrote above with change of base formula, it helps.

Choice #4/D seems the most correct, based on what I now understand about your question. I concur.

A helpful tip: when dealing with logarithms in particular, it's wise for one to be extra careful with parenthesis


Does this answer your question sufficiently I hope? :D

Answer 21

as much as a possible!