Solve for x: (2e^2x)-(7e^x)=15

let e^x = y

then you have
2y^2 - 7y - 15 = 0
(y - 5)(2y + 3) = 0
y = 5 or y = -3/2

e^x = 5
then x = ln5

or

e^x = -3/2
no solution, look at a graph of y = e^x, it is never negative

Thanks

To solve the equation (2e^2x) - (7e^x) = 15 for x, we can use a technique called substitution.

Step 1: Let's start by making a substitution. We'll set u = e^x. This substitution will simplify the equation.

Step 2: Substitute u into the equation. After the substitution, the equation becomes (2u^2) - (7u) = 15.

Step 3: Rearrange the equation to form a quadratic equation. Move 15 to the left side of the equation, and then rearrange terms to obtain 2u^2 - 7u - 15 = 0.

Step 4: Factor the quadratic equation. Find two numbers whose product is equal to the product of 2 and -15 (2 * -15 = -30) and whose sum is equal to -7. The numbers are -10 and 3. Therefore, the equation can be factored as (2u + 3)(u - 5) = 0.

Step 5: Set each factor equal to zero and solve. This gives us two equations:
a) 2u + 3 = 0
b) u - 5 = 0

For equation a), solve for u:
2u + 3 = 0
2u = -3
u = -3/2

For equation b), solve for u:
u - 5 = 0
u = 5

Step 6: Substitute back for u using the original substitution e^x = u:

For sub-equation a):
e^x = -3/2
There is no real solution for e^x = -3/2 since the exponential function e^x is always positive.

For sub-equation b):
e^x = 5
Take the natural logarithm of both sides to solve for x:
ln(e^x) = ln(5)
x = ln(5)

Therefore, the solution to the equation (2e^2x) - (7e^x) = 15 is x = ln(5).