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Alright so i have a 2^(2/x) * 4^(x/6)* ((1/8)^(1/x))^(1/6)=2^2*2^(1/3) After several tries i got it wrong. The answer is x1=3 x2=-1/5

Guest May 27, 2014

Best Answer 

 #3
avatar+19638 
+8

 

$$\boxed{2^{\frac{x}{2}} * 4^{\frac{x}{6}}* \left[(\frac{1}{8})^{\frac{1}{x}}\right]^{\frac{1}{6}}=2^2*2^{\frac{1}{3}} }\\\\
\Rightarrow 2^{\frac{x}{2}} *2^{2*\frac{x}{6}}*2^{-3*\frac{1}{x}*\frac{1}{6}}=2^{2+\frac{1}{3}}\\\\
\Rightarrow 2^{\frac{x}{2}} *2^{\frac{x}{3}}*2^{\left(-\frac{1}{2x}\right)}=2^{\frac{7}{3}}\\\\
\Rightarrow 2^{
\left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right)
}=2^{\frac{7}{3}} \quad | \quad ln\\\\
\Rightarrow {
\left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right)*\ln(2)
}=\frac{7}{3}*\ln(2)\\\\
\Rightarrow {
\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right
}=\frac{7}{3}\\\\$$

$$\\\Rightarrow {
\frac{3x}{3x}*\frac{x}{2}+\frac{2x}{2x}*\frac{x}{3}
-\frac{3}{3}*\frac{1}{2x}\right
}=\frac{7}{3}\\\\
\Rightarrow \frac{3x^2+2x^2-3}{6x}=\frac{7}{3} \quad | \quad *6x\\\\
\Rightarrow 5x^2-3=6x*\frac{7}{3}\\
\Rightarrow 5x^2-3=14x\\$$

$$\\\Rightarrow 5x^2-14x-3=0\\\\
\Rightarrow x^2-\frac{14}{5}x-\frac{3}{5}=0\\\\
\Rightarrow x_{1,2}=\frac{14}{2*5}\pm\sqrt{\frac{14*14}{(2*5)*(2*5)}+\frac{3}{5}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{14*14}{(10)*(10)}+\frac{20}{20}*\frac{3}{5}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196}{100}+\frac{60}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196+60}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{256}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\frac{16}{10}\\\\
\textcolor[rgb]{1,0,0}{x_1=}\frac{14}{10}+\frac{16}{10}=\frac{30}{10}=\textcolor[rgb]{1,0,0}{3}\\\\
\textcolor[rgb]{1,0,0}{x_2=}\frac{14}{10}-\frac{16}{10}=-\frac{2}{10}=\textcolor[rgb]{1,0,0}{-\frac{1}{5}}\\\\$$

heureka  May 28, 2014
 #1
avatar+2353 
+5

 

 

Okay, so here's what happened.

I wrote a 2/3 page answer solving the thing, after which I checked my answers.

My answers were wrong 

Then I used an equation solver to check your equation and it gave me 

$$x_1=0.716$$

$$x_2=6.28$$

Did you by any chance miswrite something in your equation?

 

Currently you have this;

$$2^{2/x} * 4^{x/6}* ((1/8)^{1/x})^{1/6}=2^2*2^{1/3}$$

reinout-g  May 27, 2014
 #2
avatar+9 
0

Oh sorry it was 2^(x/2), had to make an acount because i wasn't able do to the spamm security thing.

TreefAM  May 27, 2014
 #3
avatar+19638 
+8
Best Answer

 

$$\boxed{2^{\frac{x}{2}} * 4^{\frac{x}{6}}* \left[(\frac{1}{8})^{\frac{1}{x}}\right]^{\frac{1}{6}}=2^2*2^{\frac{1}{3}} }\\\\
\Rightarrow 2^{\frac{x}{2}} *2^{2*\frac{x}{6}}*2^{-3*\frac{1}{x}*\frac{1}{6}}=2^{2+\frac{1}{3}}\\\\
\Rightarrow 2^{\frac{x}{2}} *2^{\frac{x}{3}}*2^{\left(-\frac{1}{2x}\right)}=2^{\frac{7}{3}}\\\\
\Rightarrow 2^{
\left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right)
}=2^{\frac{7}{3}} \quad | \quad ln\\\\
\Rightarrow {
\left(\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right)*\ln(2)
}=\frac{7}{3}*\ln(2)\\\\
\Rightarrow {
\frac{x}{2}+\frac{x}{3}-\frac{1}{2x}\right
}=\frac{7}{3}\\\\$$

$$\\\Rightarrow {
\frac{3x}{3x}*\frac{x}{2}+\frac{2x}{2x}*\frac{x}{3}
-\frac{3}{3}*\frac{1}{2x}\right
}=\frac{7}{3}\\\\
\Rightarrow \frac{3x^2+2x^2-3}{6x}=\frac{7}{3} \quad | \quad *6x\\\\
\Rightarrow 5x^2-3=6x*\frac{7}{3}\\
\Rightarrow 5x^2-3=14x\\$$

$$\\\Rightarrow 5x^2-14x-3=0\\\\
\Rightarrow x^2-\frac{14}{5}x-\frac{3}{5}=0\\\\
\Rightarrow x_{1,2}=\frac{14}{2*5}\pm\sqrt{\frac{14*14}{(2*5)*(2*5)}+\frac{3}{5}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{14*14}{(10)*(10)}+\frac{20}{20}*\frac{3}{5}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196}{100}+\frac{60}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{196+60}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\sqrt{\frac{256}{100}}\\\\
\Rightarrow x_{1,2}=\frac{14}{10}\pm\frac{16}{10}\\\\
\textcolor[rgb]{1,0,0}{x_1=}\frac{14}{10}+\frac{16}{10}=\frac{30}{10}=\textcolor[rgb]{1,0,0}{3}\\\\
\textcolor[rgb]{1,0,0}{x_2=}\frac{14}{10}-\frac{16}{10}=-\frac{2}{10}=\textcolor[rgb]{1,0,0}{-\frac{1}{5}}\\\\$$

heureka  May 28, 2014

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