Solving an equation with an irrational exponent  TOPIC_SOLVED

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

Solving an equation with an irrational exponent  TOPIC_SOLVED

Postby Aqua Dragon on Sun Apr 19, 2009 7:18 pm

Hello there.

I've been having extreme difficulty with this problem:

(3/9^x)=243

I can't figure out how to isolate the x and find the answer. Granted, I already know the answer is -2, but what I really want is to find out how to reach this answer. The only thing that seemed to work so far was dividing both sides by 3, leaving 9^x=81. And from there I just kinda squared it and hoped something would happen. I'm sure there's a better way, and I would really appreciate it if someone would help me figure it out.
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Postby stapel_eliz on Sun Apr 19, 2009 7:40 pm

Aqua Dragon wrote:(3/9^x)=243

I will guess that the above means 3/(9^x) = 243 or, in LaTeX formatting:

. . . . .

I'm not sure what you mean when you refer, in your subject line, to "an irrational exponent"...?

To learn, in general, how to solve exponential equations, try here. Once you've learned the basic terms and techniques, note that:

. . . . .

. . . . .

. . . . .

Compare the powers to solve the equation. :wink:
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Re: Solving an equation with an irrational exponent

Postby Aqua Dragon on Sun Apr 19, 2009 7:47 pm

Yes, that was exactly what I meant.

I called it an irrational exponent because that's what this here textbook calls it, though oddly enough it never tells you how to find a way to solve it, since it obviously isn't the same as rational exponents.

I didn't even notice that section because I thought it would be titled something along the lines of irrational exponents. Thanks for pointing it out. I'll look through it and then try to solve some of these others. :wink:
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