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avatar+9673 

\(\sqrt[3]{(8^{2x})}=\sqrt{2^{(4x-1)}}\)

 Aug 22, 2016

Best Answer 

 #1
avatar+33661 
+15

Only true in the limit as x tends to minus infinity!

 Aug 22, 2016
 #1
avatar+33661 
+15
Best Answer

Only true in the limit as x tends to minus infinity!

Alan Aug 22, 2016
 #2
avatar+118677 
+10

Max's equation.

 

\(\begin{align*}\sqrt[3]{(8^{2x})}&=\sqrt{2^{(4x-1)}}\\ (8^{2x})^{1/3}&=(2^{(4x-1)})^{1/2}\\ ((8^{2x})^{1/3})^6&=((2^{(4x-1)})^{1/2})^6\\ (8^{2x})^2&=(2^{(4x-1)})^3\\ ((2^3)^{2x})^2&=(2^{3(4x-1)})\\ (2^3)^{4x}&=2^{3(4x-1)}\\ (2)^{12x}&=2^{3(4x-1)}\\ 12x&=3(4x-1)\\ 12x&=12x-3\\ 0&=-3\\ &\mbox{No solutions}\end{align*}\)

 

OR

\(\begin{align*}\sqrt[3]{(8^{2x})}&=\sqrt{2^{(4x-1)}}\\ 8^{2x/3}&=2^{((4x-1)/2)}\\ (2^{2x})&=2^{((4x-1)/2)}\\ 2x&=2x-0.5\\ 0&=-0.5\\ &no \;solution\end{align*}\\ \)

 

Thanks Alan.  ://

 Aug 22, 2016

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