LynnM Posts: 34, Reputation: 2 Junior Member #1 Oct 5, 2010, 03:13 PM
Exponential Equation
Can anyone help me with this question:

( [(9^{2x-1}) (3^{3x})^2] / [(27^{x+2})^4] ) = 81^3

I know that x=-19 but I keep getting -1.5 and 8. Step by step would be really appreciated. Thank you.
 Unknown008 Posts: 8,076, Reputation: 723 Uber Member #2 Oct 6, 2010, 01:59 AM

$\frac{(9^{2x-1}) (3^{3x})^2}{(27^{x+2})^4} = 81^3$

Okay, put everything to base 3.

$\frac{((3^2)^{2x-1}) (3^{3x})^2}{((3^3)^{x+2})^4} = (3^4)^3$

Now, work out the powers:

$\frac{(3^{4x-2}) (3^{6x})}{(3^{12x+24})} = (3^{12})$

Now, you know that $(a^x)(a^y) = a^{x+y}$ and $\frac{a^x}{a^y} = a^{x-y}$

Apply this:

$3^{4x-2 + 6x - 12x - 24} = 3^{12}$

Now, simplify:

$3^{-2x-26} = 3^{12}$

Equate the powers:

$-2x - 26 = 12$

This gives you:

$x = \frac{12 + 26}{-2} = -19$
 LynnM Posts: 34, Reputation: 2 Junior Member #3 Oct 6, 2010, 07:03 AM
Thanks once again! I now see where I was going wrong. After I worked out the powers, instead of adding the top powers, I was trying to multiply them. (Actually, I think I did try to add them at one point but still ended up with the wrong answer!) Oye! Lol!

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