1. ## Logarithms

Can you help with the solution of this system for ( x, y ). Thanks.

log x ( base 9) + log 8 ( base y ) = 2
log 9 ( base x ) + log y ( base 8 ) = 8/3

sahip

2. Since you have different bases, you must use a change of base formula to give all these logs the same base. I have attached a Word document showing this, because it is difficult to explain this clearly on this form.

Hope this helps! Good Luck!

3. Hello, sahip!

We will use the formula: . $\log_a(b) \:=\:\frac{1}{\log_b(a)}$

Solve: . $\begin{array}{cccc}\log_9x + \log_y8 &=&2 & {\color{blue}[1]} \\ \log_x9 + \log_8y &=& \frac{8}{3} & {\color{blue}[2]}\end{array}$

Using the formula:

. . ${\color{blue}[1]}\text{ becomes: }\;\log_9x + \frac{1}{\log_8y} \:=\:2\;\;{\color{blue}[3]}$
. . ${\color{blue}[2]}\text{ becomes: }\;\frac{1}{\log_9x} + \log_8y \:=\:\frac{8}{3}\;\;{\color{blue}[4]}$

Let: . $\begin{array}{ccc} X &=& \log_9x \\ Y &=& \log_8y \end{array}\;\;{\color{red}(a)}$

Then [3] and [4] becomes: . $\begin{array}{ccccccc}X + \dfrac{1}{Y} \:=\:2 & \Rightarrow & XY + 1 \:=\:2Y & {\color{blue}[5]} \\
\dfrac{1}{ X}+ Y \:=\:\frac{8}{3} & \Rightarrow & 1 + XY \:=\:\frac{8}{3}X & {\color{blue}[6]}\end{array}$

Equate [5] and [6]: . $2Y \:=\:\tfrac{8}{3}X \quad\Rightarrow\quad Y \:=\:\tfrac{4}{3}X\;\;{\color{blue}[7}$

Substitute into [6]: . $1 + X\left(\tfrac{4}{3}X\right) \:=\:\tfrac{8}{3}X$

. . which simplifies to: . $4X^2-8X+3\:=\:0$

. . which factors: . $(2X-1)(2X-3) \:=\:0$

. . and has roots: . $X \:=\:\tfrac{1}{2},\:\tfrac{3}{2}$

Substitute into [7]: . $Y \:=\:\tfrac{2}{3},\:2$

Back-substitute into (a).

. . $\left(\tfrac{1}{2},\,\tfrac{2}{3}\right)\!:\;\begi n{array}{ccccc}\log_9x \:=\:\tfrac{1}{2} &\Rightarrow& x \:=\:9^{\frac{1}{2}} & \Rightarrow & x \:=\:3 \\ \log_8y \:=\:\tfrac{2}{3} & \Rightarrow& y \:=\:8^{\frac{2}{3}} & \Rightarrow & y \:=\:4 \end{array}$

. . $\left(\tfrac{3}{2},\,2\right)\!:\;\begin{array}{cc ccc}\log_9x \:=\:\tfrac{3}{2} &\Rightarrow& x \:=\:9^{\frac{3}{2}} &\Rightarrow& x \:=\:27 \\
\log_8y \:=\:2 &\Rightarrow& y \:=\:8^2 &\Rightarrow& y \:=\:64 \end{array}$

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