# Thread: solve for a

1. ## solve for a

look.............................................. ..................solve for a
a a-2 a+1
3 4 25/ a-1 a+1 a-2
9 2 5

common.......................

2. Originally Posted by pickbrain
look.............................................. ..................solve for a
a a-2 a+1
3 4 25/ a-1 a+1 a-2
9 2 5

common.......................
Is this supposed to mean anything? Please write this clearly if you expect anyone to help you. Is this a matrix multiplication? Where is the equal sign?

-Dan

3. Hello, pickbrain!

Did you see what you wrote?

solve for a
a a-2 a+1
3 4 25/ a-1 a+1 a-2
9 2 5

common.......................
I think it's supposed to look like this: . $\frac{3^a\cdot4^{a-2}\cdot25^{a+1}}{9^{a-1}\cdot2^{a+1}\cdot5^{a-2}}$

The numerator is: . $3^2\cdot(2^2)^{a-2}\cdot(5^2)^{a+1} \;=\;3^2\cdot2^{2a-4}\cdot5^{2a+2}$

The denominator is: . $(3^2)^{a-1}\cdot2^{a+1}\cdot5^{a-2} \;=\;3^{2a-2}\cdot2^{a+1}\cdot5^{a-2}$

The problem becomes: . $\frac{3^a\cdot2^{2a-4}\cdot5^{2a+2}}{3^{2a-2}\cdot2^{a+1}\cdot5^{a-2}} \;=\;\frac{3^a}{3^{2a-2}}\cdot\frac{2^{2a-4}}{2^{a+1}}\cdot\frac{5^{2a+2}}{5^{a-2}}$

Divide ("subtract exponents"): . $3^{a-(2a-2)}\cdot 2^{(2a-4)-(a+1)}\cdot 5^{(2a+2)-(a-2)}$

. . . . . . . . . . . . . . . . . . . . . $= \;3^{a-2a+2}\cdot 2^{2a-4-a-1}\cdot 5^{2a+2-a+2}$

. . . . . . . . . . . . . . . . . . . . . $= \;3^{-a+2}\cdot2^{a-5}\cdot5^{a+4}$