Question

express as a sum or difference of logarithms without exponents.
*problem in post below*

Answer 1

\[\log_{c5}\sqrt{\frac{ x ^{2} }{ y ^{5}z ^{7}}} \]

Answer 2

answer asks "what is the equivalent sum or difference of logarithms?"

Answer 3

can't you try this: log a/b = loga - logb ?

Answer 4

the thing i'm confused about is the logc5. im not good with log. never have been and have no idea how to do anything with it. ive tried guides online but everything just made me more confused

Answer 5

actually c5 doesn't make sense to me either.
ignoring that we can write \(\huge{\sqrt{\frac{x^2}{y^5z^7}}=\frac{x}{y^{5/2}z^{7/2}}}\)
clear with this first step ??

Answer 6

yeah. just taking the square off the first one and dividing the bottom squares by two right?

Answer 7

so we can take the square root off

Answer 8

more specifically using this :
\(\sqrt{a^n}=a^{n/2}\)
ok, now log properties:
1st :
\(log (A/B)=log A -log B \)
so that would make
\(log x - log (y^{5/2}x^{7/2})\)
ok with 2nd step ??

Answer 9

yeah.

Answer 10

so would that be my answer?

Answer 11

another log property :
logAB = log A+log B
so
\(log(y^{5/2}z^{7/2})=log(y^{5/2})+log(z^{7/2}) \)
this is 3rd step
and 4th step is remaining.

Answer 12

so would i replace \[\log x - \log(y ^{5/2}x ^{7/2}) \]

Answer 13

with \[logx - \log(y ^{5/2}) + \log(y ^{7/2)}\]

Answer 14

\[\log(z ^{7/2}) i mean\]

Answer 15

its \(logx - \log(y ^{5/2}) - \log(y ^{7/2)}\)

Answer 16

now last step, use
log (a^n) = n log a

Answer 17

ah i see where i messed up there

Answer 18

where do i get the n from?

Answer 19

log (y^(5/2)) = (5/2) log y
ok ?
can u do the same with z ?

Answer 20

(7/2) log z

Answer 21

yup,so whats final answer?

Answer 22

logx - (5/2) log y - (7/2) log z?

Answer 23

\(log x - (5/2) log y -(7/2) log z.\)
good work :)

Answer 24

thank you

Answer 25

welcome :)