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# Logarithm

0
542
6

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

Nov 8, 2016

#5
+24979
+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

Nov 8, 2016

#1
+109520
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

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

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Nov 8, 2016
#2
+24979
+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

heureka  Nov 8, 2016
#3
+109520
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
+24979
+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

heureka  Nov 8, 2016
#5
+24979
+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

heureka  Nov 8, 2016
#6
+109520
0

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

Nov 9, 2016