# Thread: [SOLVED] Simultaneous Exponential Equations Problem

1. ## [SOLVED] Simultaneous Exponential Equations Problem

I need to find the value of x and y for the following equation:

$a^x * \frac {b^3} {b^x} * \frac {b^y} {(a^y * a^1)^2} = ab^6$

In order to find them, I need simultaneous equations but I just don't know how to get the simultaneous equations. Where can I get the second equation? Or is there any other way?

This is what I got so far(which may makes no sense):

Tried from numerators and denominators:
$x-2y+2=6$
$y-3-x=1$

Solved using substitution method:

$x=-12$
$y=-8$

Second attempt:
$x-2y+2=6$
$y-3-x=6$

Solved:
$x=-22$
$y=-13$

Thanks.

Original equation here.

2. ## Indices

Hello sk8teroy

Thanks for the link to the original question. This is:

$\frac{a^x}{b^{3-x}}\cdot \frac{b^y}{(a^{y+1})^2}=ab^6$

I suspect that the = sign should really be an $\equiv$ sign. In other words, the two sides of the equation have to be identically equal. So you need to work out what the indices of $a$ and $b$ are on the LHS (in terms of $x$ and $y$), and equate them to the indices on the RHS. That's where your two equations will come from.

So, let me start you off. I'll begin by bringing the two terms in the denominators of the LHS into the numerator, by changing the signs of their indices:

$\frac{a^x}{b^{3-x}}\cdot \frac{b^y}{(a^{y+1})^2}$

$= a^x \cdot b^{-(3-x)}\cdot b^y \cdot (a^{y+1})^{-2}$

$= a^x \cdot b^{-3+x}\cdot b^y\cdot a^{-2y-2}$

$= a^{x-2y-2}\cdot b^{-3+x+y}$

So when we equate the indices on each side, we get:

$a:x-2y-2 = 1$

$b:-3+x+y= 6$

Can you continue from here?