Solve for x.
3^(2x)-2*3^(x+5)+3^10= 0
I can see it as being factored to
(3^x - 3^5)(3^x - 3^5) = 0
so 3^x - 3^5 = 0
3^x = 3^5
x = 5
Rewrite the given equation using the substitutions x=rcos@ and y=rsin@. (@=theta)
Simplify your answer.
(2x-3)*(x^2+9)*(x+5)=0
To solve for x in the equation 3^(2x) - 2*3^(x+5) + 3^10 = 0, we can apply a technique called substitution.
Let's denote y = 3^x.
Now, we can rewrite the equation in terms of y:
y^2 - 2*3^5*y + 3^10 = 0
Next, let's solve this quadratic equation. We can do this either by factoring, completing the square, or using the quadratic formula.
Since this equation is not easily factorable, let's use the quadratic formula:
y = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -2*3^5, and c = 3^10. Substituting these values into the quadratic formula, we get:
y = (-(-2*3^5) ± sqrt((-2*3^5)^2 - 4(1)(3^10))) / (2(1))
Simplifying further:
y = (2*3^5 ± sqrt(4*3^10 - 4(1)(3^10))) / 2
y = (2*3^5 ± sqrt(4*3^10 - 12*3^10)) / 2
y = (2*3^5 ± sqrt(4 - 12)*3^10) / 2
y = (2*3^5 ± sqrt(-8)*3^10) / 2
Since we have a square root of a negative number, it means that there are no real solutions for y and, consequently, no real solutions for x.