Question 693478
You were very close to solving this.

{{{ 3^(2x)=7(3^x)-12 }}}
{{{ 3^(2x)-7(3^x)+12 = 0 }}}

This is quadratic in {{{ 3^x }}} and not {{{ x }}}
so you can factorize (using the roots you stated)
{{{ (3^x-3)(3^x-4)=0 }}}
so the solutions are {{{ 3^x=4 }}} and {{{ 3^x=3 }}}
or, by taking logs to base 10 {{{ xlog3 =log4 }}} and {{{ xlog3=log3 }}}
giving x=1.26185... and x=1 as stated.