1. jiteshmeghwal9

cmn fast

2. anonymous

$1/2\log_{2}x=1/3 \log_{2}y -1$ can you please simplfy

3. jiteshmeghwal9

$\log_{2}{x^2}=\log_{2}{(y-1)^3}$now it is clear that $x^2=(y-1)^3$so now solve the equation & gt ur answer:)

4. jiteshmeghwal9

and i use the property$\frac{1}{e}\log_{b}{a}=\log_{b}{a^e}$

5. jiteshmeghwal9

can u do it from here @aceace ????

6. anonymous

i dont really get the formula thing and i dont think i can do it...

7. jiteshmeghwal9

k! i give u the solution

8. jiteshmeghwal9

using the identity$(x-y)^3=x^3-y^3-3x^2y-3xy^2$

9. jiteshmeghwal9

so,$x^2=y^3-1^3-3y^21-3y1^2$

10. anonymous

i think the -1 may be separate because that is the wrong answer...

11. anonymous

@jiteshmeghwal9 formula should be: $\large e.Log_ba = Log_b(a)^e$

12. anonymous

what is that formula for?

13. anonymous

This is known as Power rule in Logarithms.. For example: $Log(x - 1)^2 \implies 2.Log(x-1)$ Or, $2.Log(x-1) \implies Log(x-1)^2$

14. jiteshmeghwal9

i meant if $\LARGE{\log_{a^e}{b}=1/e \log_{a}{b}}$then it will be$\LARGE{\log_{a}{b^e} }$

15. jiteshmeghwal9

so i m correct

16. campbell_st

you are asked to simplify $\frac{1}{2}\log _{2}(x) = \frac{1}{3}\log _{2}(y -1)$ multiply both sides by 3 $\log_{2}(y -1) = \frac{3}{2}\log(x)$

17. anonymous

ok.... that doesn't solve my problem...

18. campbell_st

now multiply both sides by 2 $2\log _{2}(y -1) = 3\log_{2}(x)$

19. anonymous

i think that the -1 is separate from teh log on the right hand side...

20. anonymous

that is wrong

21. anonymous

$\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1$ $\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y$ $\large \log_2(x)^\frac{1}{2} + Log_22 = Log_2(y)^\frac{1}{3}$ $\large Log_2(2\sqrt{x}) = Log_2(\sqrt[3]{y})$ $\Large 2 \times \sqrt{x} = \sqrt[3]{y}$

22. anonymous

that is what i got @waterineyes but the answer was different

23. anonymous

What is the answer tell me??

24. anonymous

it was...$64x^{3}=y^{2}$

25. anonymous

@waterineyes

26. anonymous

Yeah I got it...

27. anonymous

Before doing that I want to tell that: $largeLog_2 64 = 6$

28. anonymous

$\LARGE (2\sqrt{x})^6 = 64x^{3}$ $\LARGE (\sqrt[3]{y})^6 = y^2$ The answer you got IS correct. Just a different form.

29. anonymous

$\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1$ $\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y$ $3\log_2x + 6= 2\log_2y$ $\large \log_2x^3 + Log_264= \log_2y^2$ $64x^3 = y^2$

30. campbell_st

here is the solution $\frac{1}{2}\log_{2}(x) + log_{2}(2) = \frac{1}{3}\log_{2}(y)$ multiply every term by 6 $3\log_{2}(x) + 6\log_{2}(2) = 2\log_{2}(y)$ then $Log_{2}(2^6x^3) = y^2$ raise to the power of 2 $64x^3 = y^2$

31. anonymous

what is large LOG?

32. anonymous

what does it mean?

33. anonymous

here is a constant source of confusion since log is a function it is really best to write $$\log(x)$$ rather than $$\log x$$

34. anonymous

many (in fact maybe most) texts do not do this, but it would clear up the difference, for example, between $$\log(x+1)$$and $$\log (x) + 1$$ then gets obscured when you simply write $$\log x + 1$$

35. anonymous

i am assuming the problem is $\frac{1}{2}\log(x)=\frac{1}{3}\log(y-1)$ but it is hard to know from the way it is written

36. anonymous

ok but what is a large log? as seen above

37. anonymous

who knows?

38. anonymous

whatever it is, if the problem is the one i wrote you can multiply both sides by 6 and get $3\log(x)=2\log(y-1)$ $\log(x^3)=\log((y-1)^2)$ and so $x^3=(y-1)^2$

39. anonymous

if was something else, say $\frac{1}{2}\log(x)=\frac{1}{3}\log(y) - 1$ then proceed as @campbell_st above

40. anonymous

@aceace Large Log isn't a mathematical term. @waterineyes was trying to make the text bigger using a LaTeX markup command. Looks like a slash was missing, that's all.