Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
balkenator
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27^x^2-3x=1/3 (reads: 27 to the power of x-squared minus 3x, equals 1/3.)

so far i've gotten down to -9 plus or minus the square root of 69, over 6. I ahve done this problem and variations of the this problem three times now, and for some reason i am doing something wrong everytime with the radical...you don't have to give me the answer;I am just trying to figure out what my next step is because I am obviously messing up somewhere...

stapel_eliz
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27^x^2-3x=1/3 (reads: 27 to the power of x-squared minus 3x, equals 1/3.)
I will guess that you mean "27^(x^2 - 3x) = 1/3", which typesets as:

. . . . .$27^{x^2\, -\, 3x}\, =\, \frac{1}{3}$

I'm not sure where radicals are coming into play...?

Instead, try solving the exponential equation by the usual methods:

. . . . .$(3^3)^{x^2\, -\, 3x}\, =\, 3^{-1}$

. . . . .$3^{3x^2\, -\, 9x}\, =\, 3^{-1}$

Then equate the powers, solve the resulting quadratic equation, etc, etc.

balkenator
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### Re: can't figure out radical

It's the last part of the quadratic equation that is messing me up... that's why i said:
so far i've gotten down to -9 plus or minus the square root of 69, over 6. I have done this problem and variations of the this problem three times now, and for some reason i am doing something wrong everytime with the radical...you don't have to give me the answer;I am just trying to figure out what my next step is because I am obviously messing up somewhere...

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact: