# Math Help - Solving the Unknown for Indices

1. ## Solving the Unknown for Indices

Solve for 'x':

3^2x - 8.3^x - 9 = 0

2. Originally Posted by Ogsta101
Solve for 'x':

3^2x - 8.3^x - 9 = 0
$3^{2x} - 8 \times 3^x - 9 = 0 \Rightarrow (3^x)^2 - 8 (3^x) - 9 = 0 \Rightarrow w^2 - 8w - 9 = 0$ where $w = 3^x$.

After a bit more working:

Therefore $3^x = 9 = 3^2$ or $3^x = -1$.

One of these has no real solutions. The other has one real solution.

Therefore x = ......

3. Originally Posted by mr fantastic
$3^{2x} - 8 \times 3^x - 9 = 0 \Rightarrow (3^x)^2 - 8 (3^x) - 9 = 0 \Rightarrow w^2 - 8w - 9 = 0$ where $w = 3^x$.

After a bit more working:

Therefore $3^x = 9 = 3^2$ or $3^x = -1$.

One of these has no real solutions. The other has one real solution.

Therefore x = ......
Does x = 3???

I'm not completely sure because when I sub 3 back into the original question, it doesnt equal 0. So now I'm a little confused. But i do understand all the working...

4. Originally Posted by Ogsta101
Does x = 3???

I'm not completely sure because when I sub 3 back into the original question, it doesnt equal 0. So now I'm a little confused. But i do understand all the working...
$3^x = 3^2$. The bases are the same. What does x have to equal to make the indices the same?? Hint: NOT 3!!

5. Originally Posted by mr fantastic
$3^x = 3^2$. The bases are the same. What does x have to equal to make the indices the same?? Hint: NOT 3!!
Ugh - I did totally the wrong thing! Man I'm dumb. Ok I totally get it now!!!

Thanx sooooooooooo MUCH!!!

(x = 2 yeah?)

6. Originally Posted by Ogsta101
Ugh - I did totally the wrong thing! Man I'm dumb. Ok I totally get it now!!!

Thanx sooooooooooo MUCH!!!

(x = 2 yeah?)
Yeah!

7. how 2 solve: 4^x=32

8. $4^x = (2^2)^x$

Knowing that
$32 = 2^5$
Thus

2^{2x} = 2^5

Hence

2x = 5

x = 5/2
_________
Try to start new threads for new problems.Its better!