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 $\displaystyle x^6 + y=2^3\Longrightarrow y=\frac{8}{x^6}$
Is that the right answer?
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} $
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?