1. ## log problem

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

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

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 $\displaystyle y= x^2$ I am now trying to plug it inot the second equation $\displaystyle \log_{16}(xy) = 3\frac{1}{2}$. Can someone please show how to find x and y?
Use the base 16 with your 2nd equation:

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

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
$\displaystyle x\cdot x^8 = 3\frac{1}{2}$

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

$\displaystyle x^{16}=3\frac{1}{2}$
how do I go on from here?

7. Hello, scrible!

This is an ugly problem . . .

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

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

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

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

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

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

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

is the same as

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

xy=$\displaystyle x^3$ therefore

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

Can you solve from here?