## anonymous one year ago What is the value of

1. anonymous

$\log_{81} 3$

2. anonymous

@UsukiDoll

3. anonymous

@DecentNabeel

4. DecentNabeel

$\log _{81}\left(3\right)=\frac{1}{4}\quad \left(\mathrm{Decimal:\quad }\:0.25\right)$

5. anonymous

@satellite73

6. DecentNabeel

Happy:) @EllenJaz17

7. anonymous

how do you value that though? I have to show my work

8. anonymous

@DecentNabeel

9. DecentNabeel

$\mathrm{Rewrite\:}3\mathrm{\:\in\:power-base\:form:}\quad 3=81^{\frac{1}{4}}$ $=\log _{81}\left(81^{\frac{1}{4}}\right)$ $\mathrm{Apply\:the\:\log\:rule}:\quad \log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$ $\log _{81}\left(81^{\frac{1}{4}}\right)=\frac{1}{4}\log _{81}\left(81\right)$ $=\log _{81}\left(81\right)\frac{1}{4}$ $\mathrm{Apply\:the\:\log\:rule}:\quad \log _a\left(a\right)=1$ $\log _{81}\left(81\right)=1$ =1/4

10. DecentNabeel

ok @EllenJaz17

11. anonymous

$\log_{81}(3)=x\iff 81^x=3$ so what you are trying to do is figure out how to write 3 as a power of 81 since $$\sqrt[4]{81}=3$$ that tells you $81^{\frac{1}{4}}=3$ and you have the power you are looking for

12. anonymous

so it's not 1/4 its 3?

13. anonymous

another easier example $\log_7(49)=2$ because $$7^2=49$$

14. anonymous

no the answer is $$\frac{1}{4}$$ let me write it again

15. anonymous

ok sounds good

16. anonymous

$\huge \log_{81}(\color{red}x)=3\iff 81^{\color{red}x}=3$ you are trying to find $$\color{red}x$$

17. anonymous

you are looking for the power that you would raise 81 to to give you 3

18. anonymous

$\log_2(8)=3\iff 2^3=8$ they say the same thing

19. anonymous

you should be able to switch back and forth quickly

20. anonymous

the only tricky part of this one was recognizing that 3 is the fourth root of 81

21. anonymous

.sorry catching up was getting something to drink

22. anonymous

beer i hope

23. anonymous

haha no just water

24. anonymous

25. anonymous

I prefer straight tequila, gets the job done quick lol

26. anonymous

no beer chaser? dang!

27. anonymous

Haha nope!

28. anonymous

oops do you get that $$2^5=32$$ says the same thing as $$\log_2(32)=5$$ ?

29. anonymous

how would i apply that to find this value in my question?

30. anonymous

so if you were trying to solve $\log_2(32)=\color{red}x$ you would be trying to solve $2^{\color{red}x}=32$

31. anonymous

you have $\log_{81}(3)=\color{red}x$ you have to solve $81^{\color{red}x}=3$

32. anonymous

since $$\sqrt[4]{81}=x$$ that is the same think as saying $$81^{\frac{1}{4}}=3$$

33. anonymous

ooops i means "since $$\sqrt[4]{81}=3$$

34. anonymous

now you have the power you are looking for , that power is $$\frac{1}{4}$$ therefore since $81^{\frac{1}{4}}=3$ it is the same as saying $\log_{81}(3)=\frac{1}{4}$

35. anonymous

Ok i kinda get it lol

36. anonymous

yeah it is confusing at first a simpler example is it clear that $$10^4=10,000$$?