Solve for 'x':

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

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- Jan 27th 2008, 01:41 AMOgsta101Solving the Unknown for Indices
Solve for 'x':

3^2x - 8.3^x - 9 = 0 - Jan 27th 2008, 01:53 AMmr fantastic
$\displaystyle 3^{2x} - 8 \times 3^x - 9 = 0 \Rightarrow (3^x)^2 - 8 (3^x) - 9 = 0 \Rightarrow w^2 - 8w - 9 = 0$ where $\displaystyle w = 3^x$.

After a bit more working:

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

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

Therefore x = ...... - Jan 27th 2008, 02:05 AMOgsta101
- Jan 27th 2008, 03:08 AMmr fantastic
- Jan 27th 2008, 03:15 AMOgsta101
- Jan 27th 2008, 03:40 AMmr fantastic
- May 1st 2009, 11:21 PMAnita Wu
how 2 solve: 4^x=32(Wait)

- May 2nd 2009, 02:03 AMADARSH
$\displaystyle 4^x = (2^2)^x$

Knowing that

$\displaystyle 32 = 2^5 $

Thus

2^{2x} = 2^5

Hence

2x = 5

x = 5/2

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