Question 871834
<pre>
Hi and thanks for the help! I was wondering how to solve log base square root 3 to the 1/9=x. also one other question is
to change the exponential equation to a log... x^3=729 

If {{{log (sqrt(3), (1/9)) = x}}}, then:
       {{{(sqrt(3))^x = 1/9 }}} ---- Converting to EXPONENTIAL form
   {{{matrix(2,1, " ", (3^(1/2))^x) = 1/9}}} ---- Converting {{{sqrt(3)}}} to {{{matrix(2,1, " ", 3^(1/2))}}}
       {{{matrix(2,1, "", 3^(x/2)) = 1/9}}}
      {{{matrix(1,2, matrix(2,1, " ", 3^(x/2)), "="))}}}{{{matrix(2,1, " ", 3^(  -  2))}}}
          {{{x/2 =  -  2}}} ----- Bases are equal, and so are the exponents
          <font color = red><font size = 4><b>x</font></font></b> = 2(- 2) = <font color = red><font size = 4><b>- 4</font></font></b></pre>