Need help with logarithm equation.

**ShapeShifter** · August 30th 2011, 07:07 AM

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!~

**Siron** · August 30th 2011, 07:17 AM

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

**HallsofIvy** · August 30th 2011, 07:32 AM

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

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

**ShapeShifter** · August 30th 2011, 11:17 PM

yes , the expression is $\log_2\left(\frac{\sqrt{m^3\cdot n}}{64}\right)$

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

so the answer is (3/2)x + (1/2)y -6
Thanks for helping !

**Siron** · August 31st 2011, 12:20 AM

You're welcome and enjoy the forum!

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