Question 1177146
<br>
{{{3^(2x)=9^(2x)-1}}}<br>
Rewrite 9 as 3^2 to make the bases the same:<br>
{{{3^(2x) = (3^2)^(2x)-1}}}
{{{3^(2x) = 3^(4x)-1}}}
{{{3^(4x)-3^(2x)-1 = 0}}}<br>
This is a quadratic equation with "3^(2x)" as the "variable".  To make it easier to solve, let y=3^(2x), so that 3^(4x) = y^2:<br>
{{{y^2-y-1=0}}}<br>
That quadratic does not factor; using the quadratic formula gives us<br>
{{{y = (1+-sqrt(5))/2}}}<br>
Those values are one positive and one negative.  We can ignore the negative root, because y=3^(2x) is never negative.  So<br>
{{{y = 3^(2x) = (1+sqrt(5))/2}}}
{{{2x*log(3) = log(((1+sqrt(5))/2))}}}
{{{x = ((log((1+sqrt(5))/2)))/(2*log(3))}}}<br>
A calculator shows the value to be 0.219 to 3 decimal places.<br>