Logarithm

Heureka, could you please tell me how I can get my Latex to right align in front of the equal sign. ??

right alignment :

$\begin{array}{lcr} log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\ log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\ log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\ 2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\ \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\ log_2(x)&=&2(10-2log_2(x))\\ log_2(x)&=&20-4log_2(x)\\ 5log_2(x)&=&20\\ log_2(x)&=&4\\ 2^{log_2(x)}&=&2^4\\ x&=&16\\ \end{array}$

The code:

\begin{array}{lcr}
 log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\
  log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\
   log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\
   2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\
    \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\
     log_2(x)&=&2(10-2log_2(x))\\
      log_2(x)&=&20-4log_2(x)\\
       5log_2(x)&=&20\\
        log_2(x)&=&4\\
         2^{log_2(x)}&=&2^4\\
          x&=&16\\
\end{array}

\begin{array} {lcr}... &=&...\end{array}

First field:       l = left alignment

Second field:   c = center

Third field:         r = right alignment

Thanks Heureka, but I wanted both sides up against the equal sign.

I wanted right aligned on the left side

and left aligned on the right side.    ://

Do you know how to do that?

$\begin{array}{rcl} log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\ log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\ log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\ 2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\ \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\ log_2(x)&=&2(10-2log_2(x))\\ log_2(x)&=&20-4log_2(x)\\ 5log_2(x)&=&20\\ log_2(x)&=&4\\ 2^{log_2(x)}&=&2^4\\ x&=&16\\ \end{array}$

$$\begin{array}{rcl}  log_2(log_2(x))&=&log_2(10-2log_2(x))+1\\   log_2(log_2(x))-log_2(10-2log_2(x))&=&1 \\    log_2\frac{(log_2(x))}{(10-2log_2(x))}&=& 1\\    2^{log_2\frac{(log_2(x))}{(10-2log_2(x))}}&=&2^1\\     \frac{(log_2(x))}{(10-2log_2(x))}&=&2\\      log_2(x)&=&2(10-2log_2(x))\\       log_2(x)&=&20-4log_2(x)\\        5log_2(x)&=&20\\         log_2(x)&=&4\\          2^{log_2(x)}&=&2^4\\           x&=&16\\ \end{array}$$

\begin{array} {rcl}... &=&...\end{array}
left side:         r = right aligned
Second field:   c = center
right side:          l = left aligned