The value of $$\log_2 x$$ which satisfy $$6 - 9\log_{8}\left(\frac{4}{x}\right)^{\frac{1}{3}} - 8(\log_{256}x)^{\frac{2}{3}} - (\log_2 x^8)^{\frac{1}{3}} = 0$$ is
$$6-9\log_8\left(\frac{4}{x}\right)^{\frac{1}{3}}-8\left(\log_{256}x\right)^{\frac{2}{3}}-\left(\log_{2\ }x^8\right)^{\frac{1}{3}}=0$$
$$6-\log_24+\log_2\ x-2\left(\log_2x\right)^{\frac{2}{3}}-2\left(\log_2x\right)^{\frac{1}{3}}=0$$
Let $$\left(\log_2x\right)^{\frac{1}{3}}$$ be t.
$$4+t^3-2t^2-2t=0$$
or, $$\left(t-2\right)\left(t^2-2\right)=0$$
so, $$t=2$$ or
$$\log_2\ x=8$$
Create a FREE account and get: