We determine x such that 2x⋅82x+2=2⋅2x2+15(4x)2. 2x⋅82x+2=2⋅2x2+15(4x)2⇔2x⋅(23)2x+2=21⋅2x2+1542x⇔2x⋅26x+6=2x2+16(22)2x⇔27x+6=2x2+1624x⇔27x+6=2x2−4x+16⇔7x+6=x2−4x+16⇔x2−11x+10=0⇔(x−10)(x−1)=0⇔x=1 or x=10. ‹ Previous pageFeature exponential functions Next pageExercise 1 ›