Question

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

Answer 1

cmn fast

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

k! i give u the solution

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

@jiteshmeghwal9 formula should be:

\[\large e.Log_ba = Log_b(a)^e\]

Answer 12

what is that formula for?

Answer 13

This is known as Power rule in Logarithms..

For example:

\[Log(x - 1)^2 \implies 2.Log(x-1)\]

Or,

\[2.Log(x-1) \implies Log(x-1)^2\]

Answer 14

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

Answer 15

so i m correct

Answer 16

you are asked to simplify

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

multiply both sides by 3

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

Answer 17

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

Answer 18

now multiply both sides by 2

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

Answer 19

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

Answer 20

that is wrong

Answer 21

\[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\]

\[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\]

\[\large \log_2(x)^\frac{1}{2} + Log_22 = Log_2(y)^\frac{1}{3}\]

\[\large Log_2(2\sqrt{x}) = Log_2(\sqrt[3]{y})\]


\[\Large 2 \times \sqrt{x} = \sqrt[3]{y}\]

Answer 22

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

Answer 23

What is the answer tell me??

Answer 24

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

Answer 25

@waterineyes

Answer 26

Yeah I got it...

Answer 27

Before doing that I want to tell that:

\[largeLog_2 64 = 6\]

Answer 28

\[\LARGE (2\sqrt{x})^6 = 64x^{3}\]
\[\LARGE (\sqrt[3]{y})^6 = y^2\]

The answer you got IS correct. Just a different form.

Answer 29

\[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\]

\[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\]

\[3\log_2x + 6= 2\log_2y\]

\[\large \log_2x^3 + Log_264= \log_2y^2\]

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

Answer 30

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

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

Answer 31

what is large LOG?

Answer 32

what does it mean?

Answer 33

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

Answer 34

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 \(\log x + 1\)

Answer 35

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

Answer 36

ok but what is a large log? as seen above

Answer 37

who knows?

Answer 38

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

Answer 39

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

Answer 40

@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.