Question 327800
<pre><b>
solve for x: 

{{{ln (log x)}}}{{{""=""}}}{{{0}}}

Now we use the principle:  {{{ln(A)=C}}} is equivalent to {{{A=e^C}}}

{{{log(x)}}}{{{""=""}}}{{{e^0}}}

{{{log(x)}}}{{{""=""}}}{{{1}}}

Now we use the principle:  {{{log(A)=C}}} is equivalent to {{{A=10^C}}}
 
{{{x}}}{{{""=""}}}{{{10^1}}}

{{{x}}}{{{""=""}}}{{{10}}}

Edwin</pre>