1. ## log-system

Solve the system

$\displaystyle log_{a^{2}}(x)-log_{a^{4}}(y)=3$
$\displaystyle log_{a^{6}}(x)+log_{a^{8}}(y)=4$

2. Originally Posted by dhiab
Solve the system

$\displaystyle log_{a^{2}}(x)-log_{a^{4}}(y)=3$
$\displaystyle log_{a^{6}}(x)+log_{a^{8}}(y)=4$
Is $\displaystyle a$ a constant?

Here are some things you can do to the first equation.

$\displaystyle log_{a^{2}}(x)-log_{a^{4}}(y)=3$

$\displaystyle \Longrightarrow \frac{log_a(x)}{log_a(a^2)}-\frac{log_a(y)}{log_a(a^4)}=3$

$\displaystyle \Longrightarrow \frac{log_a(x)}{2}-\frac{log_a(y)}{4}=3$

$\displaystyle \Longrightarrow 2log_a(x)-log_a(y)=12$

$\displaystyle \Longrightarrow log_a(x^2)-log_a(y)=12$

$\displaystyle \Longrightarrow log_a\left(\frac{x^2}{y}\right)=12$

Keep in mind that $\displaystyle x>0$ and $\displaystyle y>0$.

3. Hello, dhiab!

Edit: I've corrected my blunder . . .

Solve the system:

. . $\displaystyle \log_{a^2}(x)-\log_{a^4}(y) \;=\; 3 \;\;[1]$

. . $\displaystyle \log_{a^6}(x)+\log_{a^8}(y)\:=\:4\;\;[2]$

We have:

. . $\displaystyle \frac{\ln x}{\ln a^2} \;{\color{red}-}\; \frac{\ln y}{\ln a^4} \;=\;3 \quad\Rightarrow\quad \frac{\ln x}{2\ln a} \;-\; \frac{\ln y}{4\ln a} \;=\;3 \quad\Rightarrow\quad 2\ln x \;-\; \ln y \;=\;12\ln a \;\;[1]$

. . $\displaystyle \frac{\ln x}{\ln a^6} \;+\; \frac{\ln y}{\ln a^8} \;=\;4 \quad\Rightarrow\quad \frac{\ln x}{6\ln a} \;+\; \frac{\ln y}{8\ln a} \;=\;4 \quad\Rightarrow\quad 4\ln x \;+\; 3\ln y \;=\;96\ln a \;\;[2]$

$\displaystyle \begin{array}{ccccccc}3\times [1]: & 6\ln x -3\ln y &=& 36\ln a \\ \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array}$

And we have: .$\displaystyle 10\ln x \:=\:132\ln a \quad\Rightarrow\quad \ln x \:=\:\tfrac{66}{5}\ln a \:=\:\ln\left(a^{\frac{66}{5}}\right)$

. . Therefore: .$\displaystyle \boxed{x \;=\;a^{\frac{66}{5}}}$

$\displaystyle \begin{array}{ccccc}\text{-}2\times [1]\!: & \text{-}4\ln x + 2\ln y &=& \text{-}24\ln a \\ \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array}$

And we have: .$\displaystyle 5\ln y \:=\:72\ln a \quad\Rightarrow\quad \ln y \:=\:\tfrac{72}{5}\ln a \:=\: \ln\left(a^{\frac{72}{5}}\right)$

. . Therefore: .$\displaystyle \boxed{y \;=\;a^{\frac{72}{5}}}$

4. Originally Posted by Soroban
Hello, dhiab!

It may be neater with natural logs . . . or maybe not.

We have:

. . $\displaystyle \frac{\ln x}{\ln a^2} + \frac{\ln y}{\ln a^4} \;=\;3 \quad\Rightarrow\quad \frac{\ln x}{2\ln a} + \frac{\ln y}{4\ln a} \;=\;3 \quad\Rightarrow\quad 2\ln x + \ln y \;=\;12\ln a \;\;[1]$

. . $\displaystyle \frac{\ln x}{\ln a^6} + \frac{\ln y}{\ln a^8} \:=\:4 \quad\Rightarrow\quad \frac{\ln x}{6\ln a} + \frac{\ln y}{8\ln a} \:=\:4 \quad\Rightarrow\quad 4\ln x + 3\ln y \;=\;96\ln a \;\;[2]$

$\displaystyle \begin{array}{ccccccc}\text{-}3\times [1]: & \text{-}6\ln x -3\ln y &=& \text{-}36\ln a \\$$\displaystyle \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array} And we have: .\displaystyle -2\ln x \:=\:60\ln a \quad\Rightarrow\quad \ln x \:=\:-30\ln a \:=\:\ln\left(a^{-30}\right) . . Therefore: .\displaystyle \boxed{x \;=\;a^{-30}} \displaystyle \begin{array}{ccccc}\text{-}2\times [1]\!: & \text{-}4\ln x - 2\ln y &=& \text{-}24\ln a \\$$\displaystyle \text{Add [2]:} & 4\ln x + 3\ln y &=& 96\ln a \end{array}$

And we have: .$\displaystyle \ln y \:=\:72\ln a \quad\Rightarrow\quad \ln y \:=\:\ln\left(a^{72}\right)$

. . Therefore: .$\displaystyle \boxed{y \;=\;a^{72}}$
HELLO : thank you but in first equation you have - not +