1. ## Logarithms

Solve for y in terms of x
$\displaystyle 6log_2 x + log_2 y=3\Longrightarrow log_2 x^6 + log_2 y=3$
Then I took the anti log of both sides.
Which gave me $\displaystyle x^6 + y=2^3\Longrightarrow y=\frac{8}{x^6}$

2. You need to think again.
$\displaystyle \log _2 \left( {x^6 } \right) + \log _2 \left( y \right) = \log _2 \left( {x^6 y} \right)$

3. Hello, OzzMan!

Solve for $\displaystyle y$ in terms of $\displaystyle x.$

. . $\displaystyle 6\log_2 x + \log_2 y\:=\:3\quad \Rightarrow\quad \log_2(x^6) + \log_2 y\:=\:3$

Then I took the anti log of both sides,
. . which gave me $\displaystyle x^6 + y\:=\:2^3$ . . . . no

but you got the right answer . . . How did you do that?

We have: .$\displaystyle \log_2(x^6) + \log_2(y) \;=\;3\quad\Rightarrow\quad \log_2(x^6y) \;=\;3$

Take antilogs: .$\displaystyle x^6y \;=\;2^3\quad\Rightarrow\quad x^6y \;=\;8\quad\Rightarrow\quad y \;=\;\frac{8}{x^6}$

4. $\displaystyle log_2(x^6y)=3$
$\displaystyle \Rightarrow x^6y=2^3$
$\displaystyle \Rightarrow y=\frac{8}{x^6}$

5. Is it possible to do this another way and still solve for y in terms of x? This is probably the only other way possible (unless I'm wrong):

$\displaystyle 6Log_2\;x + Log_2\;y=3$
$\displaystyle \Rightarrow Log_2(xy)=\frac{1}{2}$
$\displaystyle \Rightarrow xy=2^{\frac{1}{2}}$
$\displaystyle \Rightarrow y=\frac{2^\frac{1}{2}}{x}$

Different answer though. You think both are correct?

6. Originally Posted by OzzMan
Is it possible to do this another way and still solve for y in terms of x? This is probably the only other way possible (unless I'm wrong):

$\displaystyle 6Log_2\;x + Log_2\;y=3$
$\displaystyle \Rightarrow Log_2(xy)=\frac{1}{2}$ You didn't divide the $\displaystyle \log_2(y)$ by 6
$\displaystyle \Rightarrow xy=2^{\frac{1}{2}}$
$\displaystyle \Rightarrow y=\frac{2^\frac{1}{2}}{x}$

Different answer though. You think both are correct?
I've marked the line where you made a mistake.