Logarithms: Solving for Log(3) 25 and Finding Log(3) 75 and Log(5) 75

[craig100]

Hello there, I am wondering if you could offer some help on this questoin, I have been attempting to work through it, however cannot see exactly what route I should be taking to try and solve it;
Without using a calculator or table, show that;
2 < log(3) 25 < 3 (log to the base 3 of 25)
and hence find the ceiling of log(3) 75
Also calculate log(5) 75 in terms of log(3)5
Thankyou for any help you can provide.
Craig :)

Edit:

Sorry i forgot to mention, what i have done so far is;

Shown that 3^3 = 27 and 3^2 = 9, this is simply setting log(3)x = 3 and 2 and i get a values in which 25 lies in

9 < 25 < 27

 

[Joffe]

My logarithm knowledge is kind of sketchy but I think this is right:

x = log_3 25

3^x = 25

If x were 3 then RHS would be 27, if it were 2 then RHS would be 9, since 25 is between these values then so is x.

Q2:

I think there is a rule that goes something like this (someone please confirm):

log_x y = \frac{log_a y}{log_a x} (where a can be anything you like).

So

log_5 75 = \frac{log_3 75}{log_3 5}