1. ## Simultaneous Equation Indices

Solve simultaneously:

$\displaystyle 8^x / 4^y = 4$
$\displaystyle 11^(y-x) = 1/11$

2. Originally Posted by acevipa
Solve simultaneously:

$\displaystyle 8^x / 4^y = 4$
$\displaystyle 11^(y-x) = 1/11$
The first equation is readily simplified:

$\displaystyle \frac{8^x}{4^y} = 4 \Rightarrow 8^x = 4 \times 4^y = 4^{y + 1} \Rightarrow (2^3)^x = (2^2)^{y + 1} \Rightarrow 2^{3x} = 2^{2y + 2} \Rightarrow 3x = 2y + 2$.

The second is also readily simplified:

$\displaystyle 11^{y - x} = \frac{1}{11} = 11^{-1} \Rightarrow y - x = -1$.

Solve the two simplified equations simultaneously .....

3. Originally Posted by acevipa
Solve simultaneously:

$\displaystyle 8^x / 4^y = 4$
$\displaystyle 11^(y-x) = 1/11$
Use substitution method:

$\displaystyle 8^x / 4^y = 4~\iff~8^x = 4^y \cdot 4 = 4^{y+1}$ .... [1]

$\displaystyle 11^(y-x) = 1/11~\iff~ y = \frac1{121} + x$ .... [2]

Plug in the term for y of [2] into [1]:

$\displaystyle 8^x = 4^{\frac1{121} + x + 1} ~\iff~ 8^x = 4^{x+\frac{122}{121}} ~\iff~ 8^x = 4^x \cdot 4^{\frac{122}{121}}$ .... Now divide both sides by $\displaystyle 4^x$

$\displaystyle \frac{8^x}{4^x} = \left(\frac84\right)^x = 4^{\frac{122}{121}} ~\iff~ 2^x = (2^2)^{\frac{122}{121}}~\iff~ \boxed{2^x = 2^{\frac{244}{121}}}$

Therefore $\displaystyle x = \frac{244}{121}$ .... and

$\displaystyle y = \frac{245}{121}$

4. Originally Posted by earboth
Use substitution method:

$\displaystyle 8^x / 4^y = 4~\iff~8^x = 4^y \cdot 4 = 4^{y+1}$ .... [1]

$\displaystyle 11^(y-x) = 1/11~\iff~ y = \frac1{121} + x$ .... [2]

Plug in the term for y of [2] into [1]:

$\displaystyle 8^x = 4^{\frac1{121} + x + 1} ~\iff~ 8^x = 4^{x+\frac{122}{121}} ~\iff~ 8^x = 4^x \cdot 4^{\frac{122}{121}}$ .... Now divide both sides by $\displaystyle 4^x$

$\displaystyle \frac{8^x}{4^x} = \left(\frac84\right)^x = 4^{\frac{122}{121}} ~\iff~ 2^x = (2^2)^{\frac{122}{121}}~\iff~ \boxed{2^x = 2^{\frac{244}{121}}}$

Therefore $\displaystyle x = \frac{244}{121}$ .... and

$\displaystyle y = \frac{245}{121}$
Sorry to say, earboth but I think you've done all that work for nothing .....

If you look at the latex code used in the original post, you'll see $$11^(y-x) = 1/11$$.

I think the intended latex code for equation 2 was $$11^{y-x} = 1/11$$, to give $\displaystyle 11^{y-x} = 1/11$ .....

5. Originally Posted by mr fantastic
Sorry to say, earboth but I think you've done all that work for nothing .....
( ................. ) here are included some forbidden words

but nevertheles it was a nice training.