Answer
$x=-0.3976$
Work Step by Step
$9^x+4\times3^x-3=0,$
$3^{2x}+4\times3^x-3=0,$ Let's let $k=3^x,$
$k^2+4k-3=0,$
Solving the quadratic equation using te quadratic formula,
$k=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$
$k=\frac{-4\pm \sqrt{(4)^2-4(1)(-3)}}{2(1)}=-2\pm\sqrt{7},$
$k_1=-2-\sqrt 7<0$ or $k=-2+\sqrt 7\approx 0.6458$
Therefore,
$3^x=0.6458,$
$x\log3=\log(0.6458),$
$x=\frac{\log(0.6458)}{\log3}\approx -0.3976$
