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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

 May 27, 2014

Best Answer 

 #3
avatar+26396 
+10

 

\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\\

5x214x3=0x2145x35=0x1,2=1425±1414(25)(25)+35x1,2=1410±1414(10)(10)+202035x1,2=1410±196100+60100x1,2=1410±196+60100x1,2=1410±256100x1,2=1410±1610x1=1410+1610=3010=3x2=14101610=210=15

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 May 28, 2014
 #1
avatar+2354 
+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 

x1=0.716

x2=6.28

Did you by any chance miswrite something in your equation?

 

Currently you have this;

22/x4x/6((1/8)1/x)1/6=2221/3

 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.

 May 27, 2014
 #3
avatar+26396 
+10
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\\

5x214x3=0x2145x35=0x1,2=1425±1414(25)(25)+35x1,2=1410±1414(10)(10)+202035x1,2=1410±196100+60100x1,2=1410±196+60100x1,2=1410±256100x1,2=1410±1610x1=1410+1610=3010=3x2=14101610=210=15

heureka May 28, 2014

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