Solve for all real values of x: $\displaystyle log_2{\bigg(x^{log_2x}\bigg)} \ = \ 4$ (The first "2" that you see on the farther left is supposed to be in a subscript position and act as base 2 for the logarithm.)
Follow Math Help Forum on Facebook and Google+
$\log_2(x^{\log_2(x)})=4$ $x^{\log_2(x)} = 16$ $2^{(\log_2(x))^2} = 16$ $\left(\log_2(x)\right)^2 = 4$ $\log_2(x) = \pm 2$ $x = 2^{\pm 2} = \dfrac 1 4,~4$
Originally Posted by greg1313 Solve for all real values of x: $\displaystyle log_2{\bigg(x^{log_2x}\bigg)} \ = \ 4$ Another way:$\displaystyle log_2{\bigg(x^{log_2x}\bigg)} = \left(\log_2(x)\right)^2$