# Thread: SOlve the equation index

1. ## SOlve the equation index

Solve the equation $\displaystyle 2^{4-x} - 2^{3-x} = \frac{1}{8}$

2. Originally Posted by mastermin346
Solve the equation $\displaystyle 2^{4-x} - 2^{3-x} = \frac{1}{8}$
Since $\displaystyle 2^{4 - x} = 2 \cdot 2^{3 - x}$, the equation simplifies to $\displaystyle 2^{3-x} = \frac{1}{8}$. Try solving it now.

3. i get :

$\displaystyle 2^{3-x} = \frac{1}{8}$

$\displaystyle 2^{3-x} = 2^{-3}$

$\displaystyle 3-x = -3$

$\displaystyle x = 6$

then?

4. Originally Posted by mastermin346
i get :

$\displaystyle 2^{3-x} = \frac{1}{8}$

$\displaystyle 2^{3-x} = 2^{-3}$

$\displaystyle 3-x = -3$

$\displaystyle x = 6$

then?
Then you check to see whether or not it solves the original equation.

5. $\displaystyle \left(\frac{2^4}{2^x}\right) - \left(\frac{2^3}{2^x}\right) = \frac{1}{8}$

$\displaystyle \frac{1}{2^x}\left(16 - 8 \right) = \frac{1}{8}$

$\displaystyle \frac{8}{2^x} = \frac{1}{8}$

$\displaystyle 64 = 2^x$

$\displaystyle 2^6 = 2 ^x$

$\displaystyle x = 6$

am i right?

6. as said above by Mr F, plug x=6 in your question to check if it satisfies the equality.

7. Originally Posted by mastermin346
$\displaystyle \left(\frac{2^4}{2^x}\right) - \left(\frac{2^3}{2^x}\right) = \frac{1}{8}$

$\displaystyle \frac{1}{2^x}\left(16 - 8 \right) = \frac{1}{8}$

$\displaystyle \frac{8}{2^x} = \frac{1}{8}$

$\displaystyle 64 = 2^x$

$\displaystyle 2^6 = 2 ^x$

$\displaystyle x = 6$

am i right?
As I said before, substitute your answer into the given equation. Does it work? If it does work, then yes x = 6 is right. If it doesn't work then no x = 6 is not right.

You do not need someone to check this answer ....