## AonZ 3 years ago Last question!!! log 1/5 (25)

1. AonZ

check the identity by evaluating...

2. AonZ

$\huge \log_{\frac{ 1 }{ 5 }}25$

3. hartnn

use the same property i gave for last Q

4. ParthKohli

$$\frac{1}{5} = 5^{-1}$$. What would you multiply to that to get $$5^2$$?

5. AonZ

|dw:1360930788428:dw|

6. hartnn

yes, simplify numerator and denominator separately.

7. agent0smith

$\large \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{5} 25 }{ \log_{5}\frac{ 1 }{ 5}} = \frac{ \log_{5} 5^2 }{ \log_{5} 5^{-1}}$

8. ParthKohli

Why so serious?

9. ParthKohli

Just figure out what you multiply to $$5^{-1}$$ to get $$5^2.$$

10. AonZ

ahh @agent0smith explained it pretty good :)

11. AonZ

@agent0smith How did you get the base to be 5? Like all of them?

12. agent0smith

Not sure what you mean... I just used base 5 because it makes it easy to solve.

13. ParthKohli

@AonZ Change of bases.

14. hartnn

you can choose any base for using that property,e,5,10,... thats why its unspecified in the property

15. AonZ

so if you use change of base rule you can choose ANY base?

16. ParthKohli

Yes.

17. AonZ

wow i reckon i understand logs a lot better now :P

18. agent0smith

^ correct @ParthKohli shouldn't it be "figure out what power you need to raise 5^-1 to, to get 25" as opposed to "Just figure out what you multiply to 5^−1 to get 5^2"?

19. ParthKohli

Yes, I am sorry.

20. ParthKohli

But you get the point.

21. agent0smith

Ah okay, cos i was wondering how this helps:$5^{-1} \times 5^3 = 25$compared to this $(5^{-1})^{-2}=25$

22. ParthKohli

:-)

23. agent0smith

@AonZ you could also do it this way (but it makes it a bit more difficult): $\large \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{25} 25 }{ \log_{25}\frac{ 1 }{ 5}} = \frac{1 }{ \log_{25} 25^{-0.5}}$

24. hartnn

or simply $$\huge \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{} 5 }{ \log_{}\frac{ 1 }{ 5}}$$

25. ParthKohli

Much better up there.

26. hartnn

$$\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}$$

27. agent0smith

^ that works fine, but requires a calculator.

28. ParthKohli

Nope.

29. hartnn

$$\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}= \frac{ \log_{} 5^2 }{ \log_{}\frac{ 1 }{ 5}}$$

30. hartnn

calcy for what ?

31. ParthKohli

$\dfrac{\log _5 5^2}{\log _5 5^{-1}} =$

32. agent0smith

I'm assuming @hartnn is using base 10...

33. ParthKohli

$\dfrac{2}{-1}$

34. hartnn

does not matter what base i use, by default its 10

35. ParthKohli

You can use change of base too.

36. agent0smith

$\frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}$ I mean, how are you solving this w/o changing base again, or using a calculator?

37. hartnn

$$\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}= \frac{ \log_{} 5^2 }{ \log_{}\frac{ 1 }{ 5}}=\dfrac{2 \log 5}{-1 \log 5}=-2$$

38. agent0smith

Oh you're simplifying to $\frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}} = \frac{ 2 \log5 }{ -\log5 }$