My Math Forum Multiplication

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

August 3rd, 2013, 06:18 AM   #1
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Multiplication

I'm kinda confused =)
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August 3rd, 2013, 06:32 AM   #2
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Re: Multiplication

Hello, Drake!

Quote:
 I'm kinda confused.[color=beige] . [/color][color=blue]So am I.[/color] [color=beige]. . [/color]$\sqrt{x^3\,\cdot\,3^{\frac{1}{x^3}}\,\cdot\,\frac{ 1}{x^3}\,\cdot\,3^{x^3}}$

Are we to assume that we are to simplify the expression?

$\text{W\!e have: }\:\sqrt{\left(x^3\,\cdot\,\frac{1}{x^3}\right)\le ft(3^{\frac{1}{x^3}}\,\cdot\,3^{x^3}\right)}$

[color=beige]. . . . . . [/color]$=\;\sqrt{1\,\cdot\,3^{x^3+\frac{1}{x^3}}}$

[color=beige]. . . . . . [/color]$=\;\sqrt{3^{\frac{x^6+1}{x^3}}$

[color=beige]. . . . . . [/color]$=\;\left(3^{\frac{x^6+1}{x^3}}\right)^{\frac{1}{2} }$

[color=beige]. . . . . . [/color]$=\;3^{\frac{x^6+1}{2x^3}}$

 August 3rd, 2013, 06:47 AM #3 Member   Joined: Apr 2013 Posts: 65 Thanks: 0 Re: Multiplication Yes, exactly what I've found. However my book gives the answer 3
 August 3rd, 2013, 06:49 AM #4 Member   Joined: Apr 2013 Posts: 65 Thanks: 0 Re: Multiplication I can post the initial problem if it's needed.
August 3rd, 2013, 08:25 AM   #5
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Re: Multiplication

Hello, "Drake!

Quote:
 I can post the initial problem if it's needed.

Do you mean that what you gave us was not the inital problem?

That might explain where the "3" comes from . . .

August 3rd, 2013, 09:23 PM   #6
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Re: Multiplication

Quote:
 Originally Posted by soroban [color=beige]. . . . . . [/color]$=\;3^{\frac{x^6+1}{2x^3}}$
Well, IF answer is 3, then obvious that x=1.

 August 5th, 2013, 06:16 AM #7 Member   Joined: Apr 2013 Posts: 65 Thanks: 0 Re: Multiplication Well, yes. That was only a part of the solution. Anyways, here is the problem: $x^3\,3^{\frac{1}{x^3}}\,+\,\frac{1}{x^3}\,\,3^{x^3 }\,=\,6$ We're asked to find all real values of x satisfying this equation. To solve it, the book used AM > GM inequality. $x^3\,3^{\frac{1}{x^3}}\,+\,\frac{1}{x^3}\,\,3^{x^3 }\,\geq\, 2\sqrt{x^3\,3^{\frac{1}{x^3}}\,\frac{1}{x^3}\,\,3^ {x^3}}$ And then it says $x^3\,3^{\frac{1}{x^3}}\,\frac{1}{x^3}\,\,3^{x^3}\,=\,9$ without any further explanation And $x^3\,3^{\frac{1}{x^3}}\,+\,\frac{1}{x^3}\,\,3^{x^3 }\,\geq\, 6$ Thus x=1

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