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log2(log2(x))=log2(10-2.log2(x))+1

 Nov 8, 2016

Best Answer 

 #5
avatar+24342 
+5

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

 

laugh

 Nov 8, 2016
 #1
avatar+108608 
0

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

 

log2(log2(x))=log2(10-2.log2(x))+1

 

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

.
 Nov 8, 2016
 #2
avatar+24342 
+5

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

 

laugh

heureka  Nov 8, 2016
 #3
avatar+108608 
0

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?

 Nov 8, 2016
 #4
avatar+24342 
+5

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

 

laugh

heureka  Nov 8, 2016
 #5
avatar+24342 
+5
Best Answer

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

 

laugh

heureka  Nov 8, 2016
 #6
avatar+108608 
0

Thanks Heureka, that is exactly what I wanted. :D

 Nov 9, 2016

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