To solve the equation 3(2^(2x+3)) – 5(2^(x+2)) – 156 = 0, we can use a substitution. Let’s replace 2^x with another variable, say y. Then we have:
Let y = 2^x
Substituting this into the equation gives us:
3(2^(2x) * 2^3) – 5(2^x * 2^2) – 156 = 0
Simplifying this equation gives us:
24y^2 – 20y – 156 = 0
We can solve for y using the quadratic formula:
y = (-b ± sqrt(b^2 – 4ac)) / 2a
where a = 24, b = -20, and c = -156.
Substituting these values into the formula gives us:
y = (20 ± sqrt((-20)^2 – 4(24)(-156))) / 2(24)
y = (20 ± sqrt(400 + 14976)) / 48
y = (20 ± sqrt(15376)) / 48
y = (20 ± 124) / 48
So, we have two possible solutions for y:
y1 = 23/12 and y2 = -13/6
Now we can substitute these values back into the original equation to find the corresponding values of x.
For y1 = 23/12:
2^x = y1 = 23/12 x = log2(y1) x = log(y1) / log(2) x ≈ 1.61
For y2 = -13/6:
2^x = y2 = -13/6 This is not a valid solution since 2^x can never be negative.
Therefore, the only solution to the equation is x ≈ 1.61.