Logarithms

**OzzMan** · February 8th 2008, 11:04 AM

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

**Plato** · February 8th 2008, 11:54 AM

You need to think again.
$\log _2 \left( {x^6 } \right) + \log _2 \left( y \right) = \log _2 \left( {x^6 y} \right)<br />$

**Soroban** · February 8th 2008, 12:02 PM

Hello, OzzMan!

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

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

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

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

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

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

**OzzMan** · February 8th 2008, 12:03 PM

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

**OzzMan** · February 8th 2008, 03:33 PM

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):

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

Different answer though. You think both are correct?

**earboth** · February 9th 2008, 12:41 AM

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):

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

Different answer though. You think both are correct?

I've marked the line where you made a mistake.

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