## AonZ Group Title Log question. Is there a rule that can solve this? $\huge \log_{\frac{ 1 }{ a }} \frac{ 1 }{ n }$ one year ago one year ago

1. AonZ

@agent0smith can you help with this?

2. agent0smith

Yeah i'm not sure what you want to do with it... you can change it into a couple of different forms?

3. yakzo

There is a formula for changing the base of the logarithm.

4. AonZ

well il type full equation prove log a(x) = -log 1/2 (x)

5. hartnn

use : $$\huge \log_ab=\dfrac{\log b}{\log a}$$

6. hartnn

u sure, this is the Q ? prove log a(x) = -log 1/2 (x)

7. AonZ

wait my bad LOL log a(x) = -log 1/a (x)

8. hartnn

right, use the property i gave on RIGHT side.

9. AonZ

|dw:1360928528477:dw|

10. AonZ

|dw:1360928591721:dw| these cancel right?

11. hartnn

simplify both numerator and denominator. and - log x = log(1/x) was unnecessary and no, not directly.

12. AonZ

how else do i do it then??

13. hartnn

|dw:1360928748794:dw|

14. AonZ

oh so the negatives cancel?

15. hartnn

yesh, now negatives can cancel.

16. hartnn

again apply that property on what you have...

17. hartnn

$$\huge \dfrac{\log b}{\log a}=\log_ab$$

18. AonZ

yes i can see how the equal each other now :)

19. agent0smith

$\log_{a} x = -\log_{\frac{ 1 }{ a }} x$ Use the log rule on the right side $-\log_{\frac{ 1 }{ a }} x =- \frac{ \log_{a} x }{ \log_{a} \frac{ 1 }{ a } } =- \frac{ \log_{a} x }{ \log_{a} a ^{-1}}=- \frac{ \log_{a} x }{- \log_{a} a} = \frac{ \log_{a} x }{1}$

20. agent0smith

moved the negatives, makes it easier to follow. $-\log_{\frac{ 1 }{ a }} x = \frac{ - \log_{a} x }{ \log_{a} \frac{ 1 }{ a } } = \frac{- \log_{a} x }{ \log_{a} a ^{-1}}= \frac{- \log_{a} x }{- \log_{a} a} = \frac{ \log_{a} x }{1}$

21. AonZ

ahh ok thanks to all of you :)