Question 1114566
<br>
{{{log(2x,(48*root(3,3)))=n}}} --> {{{(2x)^n = 48*root(3,3)}}}
{{{log(3x, (162 * root(3,2)))=n}}} --> {{{(3x)^n = 162*root(3,2)}}}<br>
Then<br>
{{{(2x)^n/(3x)^n = (48*root(3,3))/(162*root(3,2))}}}
{{{(2/3)^n = ((2^4)(3^(4/3)))/((3^4)(2^(4/3)))}}}
{{{(2/3)^n = (2^(8/3))/(3^(8/3))}}}
{{{n = 8/3}}}<br>
Then solve either of the original expressions in exponential form to solve for x:<br>
{{{(2x)^(8/3) = 48*root(3,3)}}}
{{{(2)^(8/3)(x)^(8/3) = (2^4)(3^(4/3))}}}
{{{x^(8/3) = (2^(4/3))(3^(4/3)) = 6^(4/3)}}}
{{{x = sqrt(6)}}}<br>
Answer: There is one value -- {{{x=sqrt(6)}}} -- that satisfies the given equation.