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
