Log 36/5 + log 3/2 - log 6/5 ? assume all logs are to the base 3. plz answer with detail?

Answer 1

note:

log (xy) = log x + log y

and

log (x/y) = log x - log y

therefore:

log (36/5) + log (3/2) - log (6/5)

= log {[(36/5)(3/2)]/(6/5)}

= log {(36/5)(3/2)(5/6)}

= log {(3)(2)(6/5)(3)(1/2)(5/6)

= log 9

= log (3^2)

since your base is 3

and

log (a^n) to the base a = n

therefore

log (36/5) + log (3/2) - log (6/5) = 2

Answer 2

Assume all logs are to base 3 in the following:-
log [(36/5 x 3/2) / 6/5 ]
= log [(54/5) / (6/5)]
= log (9) = 2

Answer 3

Using logx + logy = log(xy) and
logx - logy = log(x/y)

we get
log((36/5)*(3/2)/(6/5))
= log((36/5)*(3/2)*(5/6))
= log(9)

Since the log base is 3, and 90 = 3^2,
the answer is 2.

Answer 4

write as log[(36/5 * 3/2)/(6/5)]
You can simplify the rest

Answer 5

log3/5 over log3/36+log3/2-log3/5 over log3/6
=(log3/5 over log3/36)*log3/2*( log3/6 over log3/5)
=(5 over 36)*2*（6 over 5)
=1 over 3

* refer to times

Answer 6

answer is 2

simplification is available at

http://www.geocities.com/regform20052007/mathpro21/1.html