Several old logarithmic and other problems

**JohnDyer** · March 27th 2012, 05:53 PM

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!

**Soroban** · March 27th 2012, 07:24 PM

Hello, JohnDyer!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**JohnDyer** · March 28th 2012, 03:30 PM

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?

**JohnDyer** · March 30th 2012, 06:30 AM

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!

Math Help - Several old logarithmic and other problems

LinkBack

Thread Tools

Display

Several old logarithmic and other problems

Re: Several old logarithmic and other problems

Re: Several old logarithmic and other problems

Re: Several old logarithmic and other problems

Similar Math Help Forum Discussions

logarithmic and exponential problems

Two Logarithmic Function Problems

Exponential and Logarithmic problems

Logarithmic problems.

logarithmic problems

Search Tags