Question 827122
<pre>
log base 16 of x + log base 4 of x + log base 2 of x = 7

solve for x

The solution, x = 2^(42/11), from the other person, is WRONG!!

   {{{log (16, (x)) + log (4, (x)) + log (2, (x)) = 7}}}
     {{{log ((x))/log ((16)) + log ((x))/log ((4)) + log (2, (x)) = 7}}}
   {{{log (2, (x))/log (2, (16)) + log (2, (x))/log (2, (4)) + log (2, (x)) = 7}}} ------ Changing from base 10 to base 2
    {{{log (2, (x))/4 + log (2, (x))/2 + log (2, (x)) = 7}}} ----- {{{matrix(1,7, log (2, (16)), and, log (2, (4)), "=", 4, and, 2)}}}, respectively
{{{log (2, (x)) + 2*log (2, (x)) + 4*log (2, (x)) = 28}}} ----- Multiplying by LCD, 4
                   {{{7*log (2, (x)) = 28}}}
                     {{{log (2, (x)) = 28/7}}}
                     {{{log (2, (x)) = 4}}}
                          {{{highlight(x = 2^4 = 16)}}}</pre>