`log_(2)` log x is meaningful if `xgt1`.
Since `4^(log_(2)logx)=2^(2log_(2)logx=(2^(log_(2)logx))^(2)`
`=(logx)^(2)(a^(log)a^(x)=x,agt0,a!=1)`
So the given equation reduces to `2(logx)^(2)-logx-1=0`.
Therefore, log x=1 or log `x=-1//2`. But `xgt1,log `xgt0`.
Hence, log x=1, i.e,x=e.
