If y = 2^x, solve the equation 8(4^x) + 9(2^x) + 1 = 0 in terms of y.

We first see that we are asked to solve the equation in terms of y. We can see that y = 2^x, and there is also a 2^x in the equation, meaning we can instantly change that to y. However, there is also a 4^x, which we derive by multiplying y by 2. Therefore, 2y = 4^x. Now that we have both expressions in terms of y, we can substitute y into the equation to get: 8(y^2) + 9(y) + 1 = 0 We then expand all of the brackets. 8y^2 + 9y +1 = 0 As we have a quadratic equation, we can factorise and solve it. (8y-1)(y-1) Therefore y= 1/8 and 1. We then refer back to the original equation and see that we need to work out the value of x for both values of y. 2^x = 1/8 x = -3

2^x = 1 x = 0

SR

Related Maths A Level answers

All answers ▸

The line L has equation y=5-2x. Find an equation of the line perpendicular to L, which passes through the point P (3,-1).


Find the gradient of a curve whose parametric equations are x=t^2/2+1 and y=t/4-1 when t=2


How do you multiply matrices together?


Given that y = 16x + 1/x , find the two values of x for which dy/dx = 0