## anonymous 5 years ago evaluate the following: log5 (1/125)

1. myininaya

[ln1/125]/[ln5]=[ln125^(-1)]/[ln5]=[ln5^(-3)]/[ln5] see if that helps

2. myininaya

what do we do next

3. myininaya

you can bring that -3 down right? does anything cancel?

4. anonymous

get rid of the exponent?

5. anonymous

oh yes i was rite. um the 5's cancel rite?

6. myininaya

so we have -3(ln5)/(ln5)=-3 the ln5 cancels

7. anonymous

so where left with -3?

8. anonymous

hello?

9. myininaya

?

10. anonymous

We can write it as $\log_{5}1-\log_{5}125$ .................. (1) then we know that logarithm of 1 with any base is zero so the (1) becomes $0-\log_{5}125$ ................... (2) now we can write this as $-\log_{5}5^{3}$ ................... (3) now this becomes $-3\times \log_{5}5$ ................... (4) and as we know $\log_{5}5=1$ ...................... (5) so answer is -3

11. anonymous

i got the rite answer :) thankyou!

12. myininaya

very good thats what we got above

13. anonymous

But in more simple way. :)