# logarithms

• Jan 7th 2007, 07:05 PM
lrn24gve
logarithms
I don't understand this problem:

solve the following system for (x;y):
log x base of 9 + log 8 base of y = 2
log 9 base of x + log y base of 8 = 8/3
• Jan 7th 2007, 07:41 PM
ThePerfectHacker
Quote:

Originally Posted by lrn24gve
I don't understand this problem:

solve the following system for (x;y):
log x base of 9 + log 8 base of y = 2
log 9 base of x + log y base of 8 = 8/3

$\displaystyle \log_9 x+\log_y 8=2$
$\displaystyle \log_x 9+\log_8 y=8/3$
Multiply the two equations,
$\displaystyle (\log_9 x+\log_y 8)(\log_x 9+\log_8 y)=2(8/3)=16/3$
$\displaystyle \log_9 x\log_x 9+\log_y 8\log_x 9+\log_8 y\log_9 x+\log_y 8\log_8 y=16/3$
$\displaystyle 2+\log_y8\log_x 9+\log_8 y\log_9 x=16/3$
$\displaystyle \log_y 8\log_x 9+\log_8 y\log_9 x=10/3$
$\displaystyle z=\log_8 y\log_9 x$
$\displaystyle 1/z+z=10/3$
$\displaystyle 3+3z^2=10z$
$\displaystyle 3z^2-10z+3=0$
$\displaystyle z=\frac{10\pm \sqrt{100-36}}{6}=1/3,3$
The first equation,
$\displaystyle \log_9 x+\frac{1}{\log_8 y}=2$
$\displaystyle A=\log_9 x, B=\log_8 y$
Thus,
$\displaystyle A+\frac{1}{B}=2$
$\displaystyle z=AB=1/3,3$
This can be solved simply.
From the A,B you can find x,y.
• Jan 7th 2007, 08:23 PM
CaptainBlack
Quote:

Originally Posted by lrn24gve
I don't understand this problem:

solve the following system for (x;y):
log x base of 9 + log 8 base of y = 2
log 9 base of x + log y base of 8 = 8/3

Convert all the logs to a common base ($\displaystyle e$ - natural logarithms will do) then we have:

$\displaystyle \frac{\ln(x)}{\ln(9)}+\frac{\ln(8)}{\ln(y)}=2$
$\displaystyle \frac{\ln(9)}{\ln(x)}+\frac{\ln(y)}{\ln(8)}=\frac{ 8}{3}$

Which is now a pair of simultaneous equations in $\displaystyle \ln(x)$ and $\displaystyle \ln(y)$ which have to be solved.

RonL
• Jan 8th 2007, 10:57 AM
Soroban
Hello, lrn24gve!

My solution is not an improvement on the others . . .

Quote:

Solve the following system for $\displaystyle (x,y):$

. . $\displaystyle \begin{array}{cc}\log_9x + \log_y8\:=\:2 \\ \log_x9+ \log_8y \:=\:\frac{8}{3}\end{array}$

Theorem: .$\displaystyle \log_ba \:=\:\frac{1}{\log_ab}$

The equations become:

. . $\displaystyle \begin{array}{cc}\log_9x + \frac{1}{\log_8y} &=\:2 \\ \frac{1}{\log_9x} + \log_8y & =\:\frac{8}{3}\end{array} \begin{array}{cc}\bf{(1)}\\ \bf{(2)}\end{array}\;\; \begin{array}{cc}\Rightarrow \\ \Rightarrow\end{array}\;\; \begin{array}{cc}(\log_9x)(\log_8y) + 1 & =\:2(\log_8y) \\ 1 + (\log_9x)(\log_8y) & =\:\frac{8}{3}(\log_8x)\end{array}$

Hence: .$\displaystyle 2(\log_8y) \:=\:\frac{8}{3}(\log_9x)\quad\Rightarrow\quad \log_8y \:=\:\frac{4}{3}(\log_9x)\;\;\bf{(3)}$

Subsitute into (1): .$\displaystyle \log_9x + \frac{1}{\frac{4}{3}(\log_9x)} \:=\:2$

. . which simplifies to: .$\displaystyle 4(\log_9x)^2 - 8(\log_9x) + 3 \:=\:0$

. . which factors: .$\displaystyle \left[2\log_9x - 1\right]\left[2\log_9x - 3\right] \:=\:0$

And we have two equations to solve:

. . $\displaystyle 2\log_9x - 1 \:=\:0\quad\Rightarrow\quad \log_9x = \frac{1}{2}\quad\Rightarrow\quad x \,=\,9^{\frac{1}{2}}\quad\Rightarrow\quad\boxed{x \,=\,3}$
. . . . Substitute into (3): .$\displaystyle \log_8y \,= \,\frac{4}{3}(\log_93) \,=\,\frac{4}{3}\cdot\frac{1}{2}\,=\,\frac{2}{3}$
. . . . . . Then: .$\displaystyle y \,= \,8^{\frac{2}{3}}\quad\Rightarrow\quad\boxed{y\,=\ ,4}$

. . $\displaystyle 2\log_9x - 3\:=\:0\quad\Rightarrow\quad \log_9x = \frac{3}{2}\quad\Rightarrow\quad x\,=\,9^{\frac{3}{2}}\quad\Rightarrow\quad\boxed{x \,=\,27}$
. . . . Substitute into (3): .$\displaystyle \log_8y \,=\,\frac{4}{3}(\log_927) \,=\,\frac{4}{3}\cdot\frac{3}{2} \,= \,2$
. . . . . . Then: .$\displaystyle y \,=\,8^2\quad\Rightarrow\quad\boxed{y \,= \,64}$

Solutions: .$\displaystyle (x,y) \:=\:(3,4),\:(27,64)$