- anonymous

hey can you guys please help with this question...

- schrodinger

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- jiteshmeghwal9

cmn fast

- anonymous

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

- 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:)

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## More answers

- jiteshmeghwal9

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

- jiteshmeghwal9

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

- anonymous

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

- jiteshmeghwal9

k! i give u the solution

- jiteshmeghwal9

using the identity\[(x-y)^3=x^3-y^3-3x^2y-3xy^2\]

- jiteshmeghwal9

so,\[x^2=y^3-1^3-3y^21-3y1^2\]

- anonymous

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

- anonymous

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

- anonymous

what is that formula for?

- anonymous

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

- jiteshmeghwal9

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

- jiteshmeghwal9

so i m correct

- 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
\[\log_{2}(y -1) = \frac{3}{2}\log(x)\]

- anonymous

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

- campbell_st

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

- anonymous

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

- anonymous

that is wrong

- anonymous

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

- anonymous

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

- anonymous

What is the answer tell me??

- anonymous

it was...\[64x^{3}=y^{2}\]

- anonymous

- anonymous

Yeah I got it...

- anonymous

Before doing that I want to tell that:
\[largeLog_2 64 = 6\]

- 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.

- anonymous

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

- 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
\[64x^3 = y^2\]

- anonymous

what is large LOG?

- anonymous

what does it mean?

- anonymous

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

- 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 \(\log x + 1\)

- 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

- anonymous

ok but what is a large log? as seen above

- anonymous

who knows?

- 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
\[x^3=(y-1)^2\]

- anonymous

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

- 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.

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