Answer:
To solve for x in the equation 16^x - 3 2^(2x) + 2 = 0, we can apply logarithms to both sides of the equation. One option is to use the natural logarithm (ln) because it helps simplify the equation:
16^x - 3 2^(2x) + 2 = 0
16^x + 2 = 3 2^(2x)
ln(16^x + 2) = ln(3 2^(2x))
ln(16^x + 2) = ln(3) + ln(2^(2x))
ln(16^x + 2) = ln(3) + 2x ln(2)
ln(16^x + 2) - 2x ln(2) = ln(3)
Now we can solve for x using numerical or graphical methods, since there is no algebraic solution for x. One possible way is to use Newton's method of approximation, which involves iterating the formula:
xn+1 = xn - f(xn) / f'(xn)
where xn is an approximation of x, f(x) is the function we want to solve for, and f'(x) is its derivative. In this case,
f(x) = 16^x + 2 - 3 2^(2x)
f'(x) = 16^x ln(16) - 12 2^(2x) ln(2)
We can start the iteration with an initial guess of x0 = 1, for example:
x1 = x0 - f(x0) / f'(x0)
= 1 - (16^1 + 2 - 3 2^(21)) / (16^1 ln(16) - 12 2^(21) ln(2))
= 0.545
x2 = x1 - f(x1) / f'(x1)
= 0.545 - (16^0.545 + 2 - 3 2^(20.545)) / (16^0.545 ln(16) - 12 2^(2*0.545) * ln(2))
= 0.678
x3 = x2 - f(x2) / f'(x2)
= 0.678 - (16^0.678 + 2 - 3 2^(20.678)) / (16^0.678 ln(16) - 12 2^(2*0.678) * ln(2))
= 0.736
x4 = x3 - f(x3) / f'(x3)
= 0.736 - (16^0.736 + 2 - 3 2^(20.736)) / (16^0.736 ln(16) - 12 2^(2*0.736) * ln(2))
= 0.764
x5 = x4 - f(x4) / f'(x4)
= 0.764 - (16^0.764 + 2 - 3 2^(20.764)) / (16^0.764 ln(16) - 12 2^(2*0.764) * ln(2))
= 0.779
We can continue the iterations until the desired level of accuracy is reached, for example, when |x5 - x4| < 0.001. In this case, the approximate solution for x is x ≈ 0.779.