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
\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\\
⇒5x2−14x−3=0⇒x2−145x−35=0⇒x1,2=142∗5±√14∗14(2∗5)∗(2∗5)+35⇒x1,2=1410±√14∗14(10)∗(10)+2020∗35⇒x1,2=1410±√196100+60100⇒x1,2=1410±√196+60100⇒x1,2=1410±√256100⇒x1,2=1410±1610x1=1410+1610=3010=3x2=1410−1610=−210=−15
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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/x∗4x/6∗((1/8)1/x)1/6=22∗21/3
Oh sorry it was 2^(x/2), had to make an acount because i wasn't able do to the spamm security thing.
\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\\
⇒5x2−14x−3=0⇒x2−145x−35=0⇒x1,2=142∗5±√14∗14(2∗5)∗(2∗5)+35⇒x1,2=1410±√14∗14(10)∗(10)+2020∗35⇒x1,2=1410±√196100+60100⇒x1,2=1410±√196+60100⇒x1,2=1410±√256100⇒x1,2=1410±1610x1=1410+1610=3010=3x2=1410−1610=−210=−15