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anonymous
 4 years ago
hey can you guys please help with this question...
anonymous
 4 years ago
hey can you guys please help with this question...

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[1/2\log_{2}x=1/3 \log_{2}y 1\] can you please simplfy

jiteshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.1\[\log_{2}{x^2}=\log_{2}{(y1)^3} \]now it is clear that \[x^2=(y1)^3\]so now solve the equation & gt ur answer:)

jiteshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.1and i use the property\[\frac{1}{e}\log_{b}{a}=\log_{b}{a^e} \]

jiteshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.1can u do it from here @aceace ????

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont really get the formula thing and i dont think i can do it...

jiteshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.1k! i give u the solution

jiteshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.1using the identity\[(xy)^3=x^3y^33x^2y3xy^2\]

jiteshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.1so,\[x^2=y^31^33y^213y1^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think the 1 may be separate because that is the wrong answer...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@jiteshmeghwal9 formula should be: \[\large e.Log_ba = Log_b(a)^e\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is that formula for?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is known as Power rule in Logarithms.. For example: \[Log(x  1)^2 \implies 2.Log(x1)\] Or, \[2.Log(x1) \implies Log(x1)^2\]

jiteshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.1i meant if \[\LARGE{\log_{a^e}{b}=1/e \log_{a}{b}}\]then it will be\[\LARGE{\log_{a}{b^e} }\]

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.2you 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok.... that doesn't solve my problem...

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.2now multiply both sides by 2 \[2\log _{2}(y 1) = 3\log_{2}(x)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think that the 1 is separate from teh log on the right hand side...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[\large \log_2(x)^\frac{1}{2} + Log_22 = Log_2(y)^\frac{1}{3}\] \[\large Log_2(2\sqrt{x}) = Log_2(\sqrt[3]{y})\] \[\Large 2 \times \sqrt{x} = \sqrt[3]{y}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is what i got @waterineyes but the answer was different

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is the answer tell me??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it was...\[64x^{3}=y^{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Before doing that I want to tell that: \[largeLog_2 64 = 6\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\LARGE (2\sqrt{x})^6 = 64x^{3}\] \[\LARGE (\sqrt[3]{y})^6 = y^2\] The answer you got IS correct. Just a different form.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[3\log_2x + 6= 2\log_2y\] \[\large \log_2x^3 + Log_264= \log_2y^2\] \[64x^3 = y^2\]

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.2here 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
 4 years ago
Best ResponseYou've already chosen the best response.0here is a constant source of confusion since log is a function it is really best to write \(\log(x)\) rather than \(\log x\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0many (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
 4 years ago
Best ResponseYou've already chosen the best response.0i am assuming the problem is \[\frac{1}{2}\log(x)=\frac{1}{3}\log(y1)\] but it is hard to know from the way it is written

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok but what is a large log? as seen above

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whatever it is, if the problem is the one i wrote you can multiply both sides by 6 and get \[3\log(x)=2\log(y1)\] \[\log(x^3)=\log((y1)^2)\] and so \[x^3=(y1)^2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if was something else, say \[\frac{1}{2}\log(x)=\frac{1}{3}\log(y)  1\] then proceed as @campbell_st above

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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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