Question

Solve algebraically, showing each step of your working, the equation (8^x-1)^2 - 18(8^x-1) + 32 = 0?

Answer 1

`x_1=log_8(3)` and `x_2=log_8(17)`

Explanation:

After using `8^x-1=y` transform, this equation became

`y^2-18y+32=0`

`(y-2)*(y-16)=0`

Hence `y_1=2` and `y2=16`

For `8^x-1=2`, `x_1=log_8(3)`

For `8^x-1=16`, `x_2=log_8(17)`