1. ## log problem

I am having trouble just trying to figureout the last part of a simultaneous equation. One part of the equation yealds $y= x^2$ I am now trying to plug it inot the second equation $\log_{16}(xy) = 3\frac{1}{2}$. Can someone please show how to find x and y?

2. It might help if you posted the original exercise...?

3. Originally Posted by scrible
I am having trouble just trying to figureout the last part of a simultaneous equation. One part of the equation yealds $y= x^2$ I am now trying to plug it inot the second equation $\log_{16}(xy) = 3\frac{1}{2}$. Can someone please show how to find x and y?
Putting your first equation into the second gives you $\log_{16}(x(x^2)) = 3\frac{1}{2}$, which simplifies to $\log_{16}(x^3) = 3\frac{1}{2}$.

Using the rule $\log(a^n) = n\log(a)$, we can rearrange the above equation to, $3\log_{16}(x) = 3\frac{1}{2}$.

Can you solve from here?

4. Originally Posted by scrible
I am having trouble just trying to figureout the last part of a simultaneous equation. One part of the equation yealds $y= x^2$ I am now trying to plug it inot the second equation $\log_{16}(xy) = 3\frac{1}{2}$. Can someone please show how to find x and y?
Use the base 16 with your 2nd equation:

$16^{\log_{16}(xy)} = 16^{\frac72}~\implies~ x \cdot y = 16^{\frac72} = 2^{14}$

Now replace y by x² and solve for x.

5. It is confusing for to solve it from there. someone help me to get to that point and I could not complete it.

6. Originally Posted by scrible
It is confusing for to solve it from there. someone help me to get to that point and I could not complete it.
Could you guide me through? This is what I did.
from you working
$x\cdot x^8 = 3\frac{1}{2}$

$
{x^2(^8)}=3\frac{1}{2}
$

$
x^{16}=3\frac{1}{2}
$

how do I go on from here?

7. Hello, scrible!

This is an ugly problem . . .

Solve the system: . $\begin{array}{cccc}y \:=\: x^2 & [1] \\ \log_{16}(xy) \:=\: \frac{7}{2} & [2] \end{array}$

[2] can be written: . $xy \:=\:16^{\frac{7}{2}} \quad\Rightarrow\quad xy \:=\:2^{14}$

Substitute [1]: . $x\cdot x^2 \:=\:2^{14} \quad\Rightarrow\quad x^3 \:=\:2^{14} \quad\Rightarrow\quad x \:=\:16\!\cdot\!2^{\frac{2}{3}}$

Substitute into [1]: . $y \:=\:\left(16\!\cdot\!2^{\frac{2}{3}}\right)^2 \:=\:256\!\cdot\!2^{\frac{4}{3}} \;=\;512\!\cdot\!2^{\frac{1}{3}}$

Therefore: . $\begin{Bmatrix} x &=& 16\sqrt[3]{4} \\ y &=& 512\sqrt[3]{2} \end{Bmatrix}$

8. I think you may be making this problem more difficult than it is.

$
\log_{16}(xy) = 3\frac{1}{2}
$

is the same as

$
xy=16^\frac{7}{2}
$

xy= $x^3$ therefore

$x^3=16^\frac{7}{2}$

Can you solve from here?