## 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 Group Title

@agent0smith can you help with this?

2. agent0smith Group Title

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

3. yakzo Group Title

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

4. AonZ Group Title

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

5. hartnn Group Title

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

6. hartnn Group Title

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

7. AonZ Group Title

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

8. hartnn Group Title

right, use the property i gave on RIGHT side.

9. AonZ Group Title

|dw:1360928528477:dw|

10. AonZ Group Title

|dw:1360928591721:dw| these cancel right?

11. hartnn Group Title

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

12. AonZ Group Title

how else do i do it then??

13. hartnn Group Title

|dw:1360928748794:dw|

14. AonZ Group Title

oh so the negatives cancel?

15. hartnn Group Title

yesh, now negatives can cancel.

16. hartnn Group Title

again apply that property on what you have...

17. hartnn Group Title

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

18. AonZ Group Title

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

19. agent0smith Group Title

$\log_{a} x = -\log_{\frac{ 1 }{ a }} x$ 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 Group Title

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 Group Title

ahh ok thanks to all of you :)