Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

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

jiteshmeghwal9Best 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:)
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.1
and i use the property\[\frac{1}{e}\log_{b}{a}=\log_{b}{a^e} \]
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.1
can u do it from here @aceace ????
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
i dont really get the formula thing and i dont think i can do it...
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.1
k! i give u the solution
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.1
using the identity\[(xy)^3=x^3y^33x^2y3xy^2\]
 one year ago

jiteshmeghwal9Best ResponseYou've already chosen the best response.1
so,\[x^2=y^31^33y^213y1^2\]
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
i think the 1 may be separate because that is the wrong answer...
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
@jiteshmeghwal9 formula should be: \[\large e.Log_ba = Log_b(a)^e\]
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
what is that formula for?
 one year ago

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

jiteshmeghwal9Best ResponseYou've already chosen the best response.1
i meant if \[\LARGE{\log_{a^e}{b}=1/e \log_{a}{b}}\]then it will be\[\LARGE{\log_{a}{b^e} }\]
 one year ago

campbell_stBest ResponseYou've already chosen the best response.2
you are asked to simplify \[\frac{1}{2}\log _{2}(x) = \frac{1}{3}\log _{2}(y 1)\] multiply both sides by 3 \[\log_{2}(y 1) = \frac{3}{2}\log(x)\]
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
ok.... that doesn't solve my problem...
 one year ago

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

aceaceBest ResponseYou've already chosen the best response.0
i think that the 1 is separate from teh log on the right hand side...
 one year ago

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

aceaceBest ResponseYou've already chosen the best response.0
that is what i got @waterineyes but the answer was different
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
What is the answer tell me??
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
it was...\[64x^{3}=y^{2}\]
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Before doing that I want to tell that: \[largeLog_2 64 = 6\]
 one year ago

WiredBest 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.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
\[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[3\log_2x + 6= 2\log_2y\] \[\large \log_2x^3 + Log_264= \log_2y^2\] \[64x^3 = y^2\]
 one year ago

campbell_stBest ResponseYou've already chosen the best response.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 \[64x^3 = y^2\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
here is a constant source of confusion since log is a function it is really best to write \(\log(x)\) rather than \(\log x\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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 \(\log x + 1\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i 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
 one year ago

aceaceBest ResponseYou've already chosen the best response.0
ok but what is a large log? as seen above
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
whatever it is, if the problem is the one i wrote you can multiply both sides by 6 and get \[3\log(x)=2\log(y1)\] \[\log(x^3)=\log((y1)^2)\] and so \[x^3=(y1)^2\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
if was something else, say \[\frac{1}{2}\log(x)=\frac{1}{3}\log(y)  1\] then proceed as @campbell_st above
 one year ago

WiredBest 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.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.