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cmn fast

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

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

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

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?

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

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

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:
\[largeLog_2 64 = 6\]

what is large LOG?

what does it mean?

ok but what is a large log? as seen above

who knows?