Given that m=2^x and n=2^y , express (log2)(Squareroot (m^3)n) / 64 in terms of x and y.

So far i got

1/2 (log2)m^3 n - (log2)2^6

1/2 (log2)m^3 n - 6

HELP!~(Doh)

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- Aug 30th 2011, 07:07 AMShapeShifterNeed help with logarithm equation.
Given that m=2^x and n=2^y , express (log2)(Squareroot (m^3)n) / 64 in terms of x and y.

So far i got

1/2 (log2)m^3 n - (log2)2^6

1/2 (log2)m^3 n - 6

HELP!~(Doh) - Aug 30th 2011, 07:17 AMSironRe: Need help with logarithm equation.
I guess the expression is:

$\displaystyle \log_2\left(\frac{\sqrt{m^3\cdot n}}{64}\right)$

What you've done already is fine:

$\displaystyle = \frac{1}{2}\cdot \log_2(m^3\cdot n) - 6$

Now use $\displaystyle \log(a\cdot b)=\log(a)+\log(b)$

And afterwards where necessary the definition to simplify:

$\displaystyle \log_a(y)=x \Leftrightarrow a^{x}=y$ - Aug 30th 2011, 07:32 AMHallsofIvyRe: Need help with logarithm equation.
I would interpret it as

$\displaystyle \log_2\left(\frac{n\sqrt{m^2}}{64}\right)$

with $\displaystyle m= 2^x$ and $\displaystyle n= 2^y$ that is

$\displaystyle \log_2\left(\frac{2^{3x/2}2^y}{2^6}\right)$

$\displaystyle \log_2(2^{3x/2})+ \log_2(2^y)- log_2(2^6)$

Quote:

What you've done already is fine:

$\displaystyle = \frac{1}{2}\cdot \log_2(m^3\cdot n) - 6$

Now use $\displaystyle \log(a\cdot b)=\log(a)+\log(b)$

And afterwards where necessary the definition to simplify:

$\displaystyle \log_a(y)=x \Leftrightarrow a^{x}=y$

- Aug 30th 2011, 11:17 PMShapeShifterRe: Need help with logarithm equation.
yes , the expression is http://latex.codecogs.com/png.latex?...n}}{64}\right)

(log2) 2^3/2x + (log2)2^1/2y - 6

so the answer is (3/2)x + (1/2)y -6

Thanks for helping ! (Clapping) - Aug 31st 2011, 12:20 AMSironRe: Need help with logarithm equation.
You're welcome and enjoy the forum! :)