1/2log[base64] - log[basex]=log6

1/2log[base64] - log[basex]=log4

\frac 12 log_{64}(?) - log_x(?)=log_{10}(6)

\frac 12 log_{64}(?) - log_x(?)=log_{10}(4)

I think you're missing something in your equations. At least you are if I understood what you wrote.

Woops! 64 and x are not bases 1/2log64-logx=log 6
1/2log64-logx=log4

Woops! 64 and x are not bases 1/2log64-logx=log 6
1/2log64-logx=log4

\frac 12 log(64) - log(x)=log(6)

\frac 12 log(64) - log(x)=log(4)

1. Recognize that a\,log(b)=log(b^a).

2. Recognize that \Large a^{\frac 12} = \sqrt{a\,}

3. Isolate the log(x) term on one side of the equation.

4. Take the antilog antilog(log(x))=10^{log(x)}=x and solve for x.