Choose any real solutions for x: log_(3)(x^2+4x)=log_(3)(35+2x) -5 7 5 No solution -7

To solve the equation $log_{3}(x^{2}+4x)=log_{3}(35+2x)$, we can use the property that if $log_{b}(x)=log_{b}(y)$, then $x=y$.<br /><br />So, we have $x^{2}+4x=35+2x$.<br /><br />Rearranging the terms, we get $x^{2}+2x-35=0$.<br /><br />Factoring the quadratic equation, we have $(x+7)(x-5)=0$.<br /><br />Setting each factor equal to zero, we get $x+7=0$ or $x-5=0$.<br /><br />Solving for $x$, we get $x=-7$ or $x=5$.<br /><br />Therefore, the real solutions for $x$ are $-7$ and $5$.