+0  
 
0
163
6
avatar

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

Guest Nov 8, 2016

Best Answer 

 #5
avatar+18956 
+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

heureka  Nov 8, 2016
Sort: 

6+0 Answers

 #1
avatar+91793 
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

 

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

Melody  Nov 8, 2016
 #2
avatar+18956 
+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+91793 
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?

Melody  Nov 8, 2016
 #4
avatar+18956 
+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+18956 
+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+91793 
0

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

Melody  Nov 9, 2016

9 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details