Thread: Several old logarithmic and other problems

I need to correct a test that I failed. Most questions I can figure out by now, but these ones I can't. They are supposed to be Algebra review, but apparently I didn't pay enough attention. Can I get an explanation on how to do these problems?
1. Solve:
log xy = 5/log z
log yz = 8/log x
log zx = 9/log y

2. Find all pairs of integers such that x^3 + y^3 = 2011

3. Find all integers x where x^2-x+1 divides x^2012+x+2001

4. f(x) is the digits of x reversed. ex: f(123) = 321
Find all 3 digit numbers where f^2(x)-x^2 is the cube of a positive integer.

5. 2x^2+3xy+2x^2 is less than or equal to 7
max(2x+y, x+2y) is less than or equal to 4.
x and y must be real.




This is my first post on this forum, I hope I didn't break any rules. Our teacher does permit external help for test corrections.

Thanks!

Hello, JohnDyer!

$\displaystyle \text{1. Solve: }\;\begin{Bmatrix}\log xy &=& \dfrac{5}{\log z} \\ \\[-3mm] \log yz &=& \dfrac{8}{\log x} \\ \\[-3mm] \log xz &=& \dfrac{9}{\log y} \end{Bmatrix}$

We have: .$\displaystyle \begin{array}{cccccccccc} \log z\log xy &=& 5 \\ \log x \log yz &=& 8 \\ \log y \log xz &=& 9 \end{array} \quad\Rightarrow\quad \begin{array}{ccccccc}\log z(\log x + \log y) &=& 5 \\ \log x(\log y+\log z) &=& 8 \\ \log y(\log x+\log z) &=& 9 \end{array}$

. . . . . . . . . . $\displaystyle \begin{array}{cccccc} \log x\log z + \log y\log z &=& 5 \\ \log x\log y + \log x\log z &=& 8 \\ \log x\log y + \log y\log z &=& 9 \end{array}$


$\displaystyle \text{Let: }\begin{Bmatrix}X &=& \log x \\ Y &=& \log y \\ Z &=& \log z\end{Bmatrix}$

$\displaystyle \text{And we have: }\; \begin{Bmatrix}&& YZ & + & XZ &=& 5 & [1] \\ XY &&& + & XZ & = & 8 & [2] \\ XY & + & YZ &&& = & 9 & [3] \end{Bmatrix}$

$\displaystyle \begin{array}{cccccc}\text{Subtract [2] - [1]:} & XY - YZ &=& 3 \\ \text{Add [3]:} & XY + YZ &=& 9 \end{array}$

. . . . . . . . . . . . . . . $\displaystyle 2XY \:=\:12 \quad\Rightarrow\quad XY \:=\:6\;\;[4]$

$\displaystyle \text{Substitute into [2]: }\:6 + XZ \:=\:8 \quad\Rightarrow\quad XZ \:=\:2\;\;[5]$

$\displaystyle \text{Substitute into [3]: }\:6 + YZ \:=\:9 \quad\Rightarrow\quad YZ \:=\:3\;\;[6] $

$\displaystyle \text{Divide [4] by [5]: }\:\frac{XY}{XZ} \,=\,\frac{6}{2} \quad\Rightarrow\quad Y \,=\,3Z$

$\displaystyle \text{Substitute into [6]: }\:(3Z)(Z) \,=\,3 \quad\Rightarrow\quad Z^2 \,=\,1 \quad\Rightarrow\quad Z \,=\,\pm1$

$\displaystyle \text{Substitute into [5]: }\:X(\pm1) \,=\,2 \quad\Rightarrow\quad X \,=\,\pm 2$

$\displaystyle \text{Substitute into [4]: }\:(\pm2)Y \,=\,6 \quad\Rightarrow\quad Y \,=\,\pm 3$

$\displaystyle \text{Back-substitute: }\:\begin{Bmatrix}\log x &=& \pm 2 \\ \log y &=& \pm 3 \\ \log z &=& \pm 1 \end{Bmatrix}$

$\displaystyle \text{Assuming base-ten, we have: }\:\begin{Bmatrix}x &=& 10^{\pm2} \\ y &=& 10^{\pm3} \\ z &=& 10^{\pm1} \end{Bmatrix}$


$\displaystyle \text{And we have two solutions: }\:(x,y,z) \;=\;(100,\,1000,\,10)\,\text{ and }\,\left(\tfrac{1}{100},\,\tfrac{1}{1000},\,\tfrac {1}{10}\right)$

Thank you so much Soroban! That was more difficult than I expected, I don't think I would have figured it out myself.

Any help on the other question?

Sorry for bumping so soon, but I need this by tonight if possible.
Yes, I do realize that nobody on this forum is paid to help, but it would be nice to have this done. These 20 point will make or break my quarterly grade.

Thanks!