simplify: sqrt[3]/(2+sqrt[3]); log_2(16sqrt[2]), etc.

dericteong

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Oct 17, 2006
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5
I'm having problem with these questions for my assignment.
Pls help...

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Hello, dericteong!

You did #1 correctly . . . Here's another approach.


\(\displaystyle \L1)\;\frac{(x^2y^2)^{-1}}{x^{-2}y^{-2}} \;=\;\frac{x^{-2}y^{-2}}{x^{-2}\,+\,y^{-2}}\)

Multiply top and bottom by \(\displaystyle x^2y^2:\)

. . \(\displaystyle \L\frac{x^2y^2}{x^2y^2}\,\cdot\frac{x^{-2}y^{-2}}{x^{-2}\,+\,y^{-2}} \;=\;\frac{x^2y^2\cdot x^{-2}y^{-2}}{x^2y^2\cdot x^{-2}\,+\,x^2y^2\cdot y^{-2}} \;=\;\frac{1}{y^2\,+\,x^2}\)

\(\displaystyle 2)\;(\sqrt{3}\,-\,5)(2\sqrt{3}\,+\,1)\)

Do the "FOIL" . . .

. . . . \(\displaystyle (\sqrt{3})(2\sqrt{3})\,+\,(\sqrt{3})(1)\,+\,(-5)(2\sqrt{3})\,+\,(-5)(1)\)

. . \(\displaystyle =\:6\,+\,\sqrt{3}\,-\,10\sqrt{3}\,-\,5\)

. . \(\displaystyle = \:1\,-\,7\sqrt{3}\)



\(\displaystyle 3)\:\L\frac{\sqrt{3}}{2\,+\,\sqrt{3}}\)

Multiply top and bottom by the conjugate of the denominator.

. . \(\displaystyle \L\frac{\sqrt{3}}{2\,+\,\sqrt{3}}\,\cdot\,\frac{2\,-\,\sqrt{3}}{2\,-\,\sqrt{3}} \:=\:\frac{2\sqrt{3}\,-\,3}{4\,-\,2\sqrt{3}\,+\,2\sqrt{3}\,-\,3}\:=\:\frac{2\sqrt{3}\,-\,3}{1}\:=\:2\sqrt{3}\,-\,3\)



\(\displaystyle 4)\;\log_2(16\sqrt{2})\)

We have: \(\displaystyle \,\log_2\left(2^4\cdot2^{\frac{1}{2}}\right) \:=\:\log_2\left(2^{\frac{9}{2}}\right) \:=\:\frac{9}{2}\)



\(\displaystyle 5)\;1\,-\,\frac{1}{2}\cdot\log_7(49)\,+\,3\cdot\log_7\left(\frac{1}{7}\right)\)

We have: \(\displaystyle \,1\,-\,\log_7\left(49^{\frac{1}{2}}\right) \,+\,\log_7\left(\frac{1}{7}\right)^3\;=\;1\,-\,\log_7(7)\,+\,\log_7\left(7^{-1}\right)^3\)

and then:. \(\displaystyle 1 \,- \,\underbrace{\log_7(7)}_\downarrow\,+\,\underbrace{\log_7(7^{-3}}_\downarrow)\)
. . . . . . \(\displaystyle = \:1\;\:-\;\,1\;\;\;+\;\;\;(-3) \;\;= \;\;-3\)

 
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