Logarithmic equations

pankaj · #1 $Old$ 08-18-2008, 04:19 AM

If $(x_{1},y_{1})$ is solution of $log_{225}(x)+log_{64}(y)=4$ and $(x_{2},y_{2})$ is solution of
$log_{x}(225)-log_{y}(64)=1$
then find the value of $log_{30}(x_{1}x_{2}y_{1}y_{2})$

I got the answer as 12.But i think that language of this question is not proper .It should have been
If $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are solutions of $log_{225}(x)+log_{64}(y)=4$ and
$log_{x}(225)-log_{y}(64)=1$
then find the value of $log_{30}(x_{1}x_{2}y_{1}y_{2})$ .
What do you guys think?
According to the answer i obtained i treated the 2 equations to be simultaneous equations in x and y and obtained 2 solutions.But according to language of the original question if $(x_{1},y_{1})$ is solution of first equation then it need not be solution of the second equation.

Soroban · #2 $Old$ 08-18-2008, 03:42 PM

Hello, pankaj!

If that's the original wording of the problem, it's criminally sloppy!

I tried two different interpretations ... and one seems to work out.

Here's one interpretation . . .

Quote:

If $(x_1,y_1)$ is any solution of $\log_{225}(x)+\log_{64}(y)\:=\:4\;\;{\color{blue}[1]}$

and $(x_2,y_2)$ is any solution of $\log_{x}(225)-\log_{y}(64)\:=\:1\;\;{\color{blue}[2]}$

then find the value of $\log_{30}(x_1x_2y_1y_2)$

A solution of [1] is: . $x_1 = 225^2,\;y_1 = 64^2$

A solution of [2] is: . $x_2 = 15,\;y_2 =64$

Then: . $x_1x_2y_1y_2 \;=\;225^2\cdot15\cdot64^2\cdot64 \;=\;(3^4\cdot5^4)\cdot(3\cdot5)\cdot(2^{12})(2^6)$

. . . . $= \;2^{18}\cdot3^5\cdot5^5 \;=\;2^{13}\cdot(2^5\cdot3^5\cdot5^5) \;-=\;2^{13}\cdot(2\cdot3\cdot5)^5 \;=\;2^{13}\cdot30^5$

Therefore: . $\log_{30}(x_1x_2y_1y_2) \;=\;\log_{30}\left(2^{13}\cdot30^5\right) \;=\;\boxed{13\log_{30}(2) + 5}$

But, of course, this result is not a constant.
It varies, depending on the solutions we select: $(x_1,y_1),\;(x_2,y_2)$

Here's another interpretation . . . and I got your answer!

Quote:

$(x_1,y_1)$ and $(x_2,y_2)$ are solutions of the system: . $\begin{array}{cccc}\log_{225}(x)+\log_{64}(y) &=& 4\\ \log_{x}(225)-\log_{y}(64) &=& 1 \end{array}$
Find the value of $\log_{30}(x_1x_2y_1y_2)$

We have: . $\begin{array}{cccc}\log_{225}(x) + \log_{64}(y) &=& 4 \\ \\[-3mm] \dfrac{1}{\log_{225}(x)} + \dfrac{1}{\log_{64}(y)} &=& 1 \end{array}$

Let: . $\begin{Bmatrix}X &=& \log_{225}(x) \\ Y &=& \log_{64}(y) \end{Bmatrix}\quad \text{ and we have: }\;\begin{array}{cccc} X + Y &=& 4 & {\color{blue}[1]} \\ \\[-4mm] \dfrac{1}{X} - \dfrac{1}{Y} &=& 1 & {\color{blue}[2]} \end{array}$

$\begin{array}{cccc}\text{{\color{blue}[2]} becomes:} & Y - X &=& XY \\ \text{Add {\color{blue}[1]}:} & Y + X &=&4 \end{array}$

And we get: . $2Y \:=\:XY + 4 \quad\Rightarrow\quad Y \:=\:\frac{4}{2-X}$

Substitute into [1]: . $X + \frac{4}{2-X} \:=\:4\quad\Rightarrow\quad X^2 - 6X + 4 \:=\:0$

The Quadratic Formula gives us: . $X \:=\:3 \pm\sqrt{5}$

Substitute into [1]: . $Y \:=\:1 \mp \sqrt{5}$

Then we have: . $\begin{array}{ccccccc}x_1 &=&225^{3+\sqrt{5}} & \quad & y_1 &=& 64^{1-\sqrt{5}} \\ x_2 &=& 225^{3-\sqrt{5}} & \quad & y_2 &=& 64^{1+\sqrt{5}} \end{array}$

Then: . $x_1x_2y_1y_2 \;=\;\left(225^{3+\sqrt{5}}\right)\left(225^{3-\sqrt{5}}\right)\left(64^{1-\sqrt{5}}\right)\left(64^{1+\sqrt{5}}\right) \;=\;225^6\cdot64^2$

. . . . . . $= \;3^{12}\cdot5^{12}\cdot2^{12} \;=\;(2\cdot3\cdot5)^{12} \;=\;30^{12}$

Therefore: . $\log_{30}\left(30^{12}\right) \;=\;\boxed{12}$

pankaj · #3 $Old$ 08-19-2008, 09:06 AM

Thanks for the verification.