I meant e^(2x)+10e^(x)-75=0.
Is it 2x^2 + 10x - 75 = 0
2(x+15)(x-5)= 0
x = -15 x = 5
or is it something else?
No it is e^(2x)+10e^(x)-75=0. It has something to do with pre-calc if that helps.
To solve the equation e^(2x) + 10e^(x) - 75 = 0, we can use a substitution technique. Let's substitute a variable y = e^x.
Substituting y = e^x into the equation, we get y^2 + 10y - 75 = 0.
Now, we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. To find the factors of the equation, we need to find two numbers whose sum is 10 and whose product is -75.
The numbers that satisfy these conditions are 15 and -5. Therefore, we can factor the equation as (y + 15)(y - 5) = 0.
Setting each factor equal to zero, we have two possible values for y: y + 15 = 0 and y - 5 = 0.
Solving for y, we get y = -15 and y = 5.
Now, we need to determine the values of x using the relationship y = e^x.
For y = -15, there is no real solution since e^x cannot be negative.
For y = 5, we can solve for x by taking the natural logarithm (ln) of both sides of the equation:
ln(5) = x.
So the solution to the equation e^(2x) + 10e^(x) - 75 = 0 is x = ln(5).