Question

log5(2) - log5(7)

Answer 1

\[\boxed{\log_b x-\log_by=\log_b\frac xy}\]

Answer 2

oh i see. let me plug them in.

Answer 3

log5(2/7) ?

Answer 4

Thats right,
now you might want to change the base,
\[\boxed{\log_b x=\dfrac{\log_c x}{ \log_c b}}\]

Answer 5

if your calculator can take logs to base 10 or logs to base e,
the choose c=10 , or c=e

Answer 6

ahh i'm not sure how to do that :/

Answer 7

What exactly was the original question?

Answer 8

you actually already helped me answer it. but i have another question

Answer 9

ok

Answer 10

what if I was adding: log5(2) + log5(7)

Answer 11

would I then multiply 2 and 7?

Answer 12

exactly;
\[\large\boxed{\log_b x+\log_by=\log_bxy}\]

Answer 13

I see. one more question!

Answer 14

log8(25)

Answer 15

is that equal to 2log8(1/5) ?

Answer 16

\[\log_8(25)=\log_8(5^2) \]
now use
\[\large\boxed{\log_bx^n=n\log_b x}\]