# Need help with logarithm equation.

• Aug 30th 2011, 08:07 AM
ShapeShifter
Need 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, 08:17 AM
Siron
Re: Need help with logarithm equation.
I guess the expression is:
$\log_2\left(\frac{\sqrt{m^3\cdot n}}{64}\right)$

What you've done already is fine:
$= \frac{1}{2}\cdot \log_2(m^3\cdot n) - 6$

Now use $\log(a\cdot b)=\log(a)+\log(b)$
And afterwards where necessary the definition to simplify:
$\log_a(y)=x \Leftrightarrow a^{x}=y$
• Aug 30th 2011, 08:32 AM
HallsofIvy
Re: Need help with logarithm equation.
Quote:

Originally Posted by Siron
I guess the expression is:
$\log_2\left(\frac{\sqrt{m^3\cdot n}}{64}\right)$

I would interpret it as
$\log_2\left(\frac{n\sqrt{m^2}}{64}\right)$
with $m= 2^x$ and $n= 2^y$ that is
$\log_2\left(\frac{2^{3x/2}2^y}{2^6}\right)$
$\log_2(2^{3x/2})+ \log_2(2^y)- log_2(2^6)$

Quote:

What you've done already is fine:
$= \frac{1}{2}\cdot \log_2(m^3\cdot n) - 6$

Now use $\log(a\cdot b)=\log(a)+\log(b)$
And afterwards where necessary the definition to simplify:
$\log_a(y)=x \Leftrightarrow a^{x}=y$
• Aug 31st 2011, 12:17 AM
ShapeShifter
Re: 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, 01:20 AM
Siron
Re: Need help with logarithm equation.
You're welcome and enjoy the forum! :)