Question

6log(c )z-1/2log(c ) x+log(c )w

Answer 1

each log has a base of c?

Answer 2

i will draw it

Answer 3

ok

Answer 4

you can always use latex or the equation editor =] @c_lapham

Answer 5

says to write as single logarithm

Answer 6

Is it:\[\bf 6\log _{c}(z)-\frac{ 1 }{ 2 }\log_c(x)+\log_c(w)\]@c_lapham

Answer 7

@c_lapham

Answer 8

yes but there isn't parenthesis on the variables if that makes a difference

Answer 9

Notice that the coefficients in front of the logarithms can be made exponents. So we can re-write the expression like this:\[\bf 6\log_c(z)-\frac{1}{2}\log_c(x)+\log_c(w)=\log_c(z)^6-\log_c(x)^{1/2}+\log_c(w)\]

Answer 10

ok

Answer 11

Now use the following rules to write as a single expression.\[\bf \log_c(a \times b)=\log_c(a)+\log(b)\]\[\bf \log_c(a \div b)=\log_c(a)-\log_c(b)\]Then you'll get the answer.
@c_lapham

I have to leave bye.

Answer 12

ok thanks for the help

Answer 13

\[ \log_{c}(z)^{6}\div (x)^{1/2 }\]

Answer 14

would this be right for the first half?

Answer 15

correct

Answer 16

what about the last log set what would i do with that

Answer 17

well you now have something like this

log(....) + log(....)

or something like this

log(A) + log(B)

so you would combine the two logs to get log(A*B)

Answer 18

but isnt there only one log left to simplify

Answer 19

doing that will give you

\[\large \log_{c}\left(\frac{z^6}{x^{1/2}}*w\right)\]

which can be rewritten to get

\[\large \log_{c}\left(\frac{wz^6}{x^{1/2}}\right)\]

Answer 20

oh i see what you did

Answer 21

\[\large \log_{c}\left(\frac{z^6}{x^{1/2}}\right) + \log_{c}\left(w\right)\]

would turn into

\[\large \log_{c}\left(\frac{z^6}{x^{1/2}}*w\right)\]

after using the rule: log(A)+log(B) = log(A*B)

Answer 22

thank you i actually understand it now

Answer 23

you're welcome