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

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cmn fast
\[1/2\log_{2}x=1/3 \log_{2}y -1\] can you please simplfy
\[\log_{2}{x^2}=\log_{2}{(y-1)^3} \]now it is clear that \[x^2=(y-1)^3\]so now solve the equation & gt ur answer:)

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Other answers:

and i use the property\[\frac{1}{e}\log_{b}{a}=\log_{b}{a^e} \]
can u do it from here @aceace ????
i dont really get the formula thing and i dont think i can do it...
k! i give u the solution
using the identity\[(x-y)^3=x^3-y^3-3x^2y-3xy^2\]
i think the -1 may be separate because that is the wrong answer...
@jiteshmeghwal9 formula should be: \[\large e.Log_ba = Log_b(a)^e\]
what is that formula for?
This is known as Power rule in Logarithms.. For example: \[Log(x - 1)^2 \implies 2.Log(x-1)\] Or, \[2.Log(x-1) \implies Log(x-1)^2\]
i meant if \[\LARGE{\log_{a^e}{b}=1/e \log_{a}{b}}\]then it will be\[\LARGE{\log_{a}{b^e} }\]
so i m correct
you are asked to simplify \[\frac{1}{2}\log _{2}(x) = \frac{1}{3}\log _{2}(y -1)\] multiply both sides by 3 \[\log_{2}(y -1) = \frac{3}{2}\log(x)\]
ok.... that doesn't solve my problem...
now multiply both sides by 2 \[2\log _{2}(y -1) = 3\log_{2}(x)\]
i think that the -1 is separate from teh log on the right hand side...
that is wrong
\[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[\large \log_2(x)^\frac{1}{2} + Log_22 = Log_2(y)^\frac{1}{3}\] \[\large Log_2(2\sqrt{x}) = Log_2(\sqrt[3]{y})\] \[\Large 2 \times \sqrt{x} = \sqrt[3]{y}\]
that is what i got @waterineyes but the answer was different
What is the answer tell me??
it was...\[64x^{3}=y^{2}\]
Yeah I got it...
Before doing that I want to tell that: \[largeLog_2 64 = 6\]
\[\LARGE (2\sqrt{x})^6 = 64x^{3}\] \[\LARGE (\sqrt[3]{y})^6 = y^2\] The answer you got IS correct. Just a different form.
\[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[3\log_2x + 6= 2\log_2y\] \[\large \log_2x^3 + Log_264= \log_2y^2\] \[64x^3 = y^2\]
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 \[64x^3 = y^2\]
what is large LOG?
what does it mean?
here is a constant source of confusion since log is a function it is really best to write \(\log(x)\) rather than \(\log x\)
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 \(\log x + 1\)
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
ok but what is a large log? as seen above
who knows?
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 \[x^3=(y-1)^2\]
if was something else, say \[\frac{1}{2}\log(x)=\frac{1}{3}\log(y) - 1\] then proceed as @campbell_st above
@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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