how to solve logx4+logx16+logx64=12 ? ,x is in the base in this question?

I ran your question logx4+logx16+logx64=12 on Wolfram Alpha and here's the output it computed:

http://www48.wolframalpha.com/input/?i=logx4%2Blogx16%2Blogx64%3D12

log_x(4) + log_x(16) + log_x(64) = 12
log_x(4 * 16 * 64) = 12
log_x(4096) = 12
x^12 = 4096
x^12 = 2^12
x = 2

Let log base = 2

Check
2 + 4 + 6 = 12 as required

Thus x = 2

logx4+logx16+logx64=12
logx(2^2+4+6)=12
logx2^12=12
12logx2=12
logx2=1
x=2

logx4+logx16+logx64=12
=>logx(4*16*64)=12
=>logx(2^2 * 2^4 * 2^6)=12
=>logx(2^12)=12
=>x^12=2^12
=>x=2

All logs are base x.

log(4) + log(16) + log(64) = 12
log(4) + log(4^2) + log(4^3) = 12
log(4) + 2log(4) + 3log(4) = 12
6log(4) = 12
log(4) = 2

Exponentiate.

4 = x^2
x = 2