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!~
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:
$\displaystyle \log_a(y)=x \Leftrightarrow a^{x}=y$
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
$\displaystyle \log_2\left(\frac{2^{3x/2}2^y}{2^6}\right)$
$\displaystyle \log_2(2^{3x/2})+ \log_2(2^y)- log_2(2^6)$
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:
$\displaystyle \log_a(y)=x \Leftrightarrow a^{x}=y$