Solve logarithmic equation (e^2x)-(7e^x)+10=0?

Question

Solve logarithmic equation (e^2x)-(7e^x)+10=0?

Answer 1

Suppose that e^x=a, we get:

a^2-7a+10=0

(a-5)(a-2)=0

a=5 or a=2

e^x=5 or e^x=2

x=ln 5 or x= ln 2

Answer 2

Let y = e^x :-

y² - 7y + 10 = 0

(y - 5)(y - 2) = 0

y = 5 , y = 2

e^x = 5 , e^x = 2

x ln e = ln 5 , x ln e = 2

x = ln 5 , x = 2

Answer 3

(e^2x)-(7e^x)+10=0

Put e^x = t, then

t^2 - 7t +10 =0

t = (7/2) +/- sqrt[49/4 - 10]

= (7/2) +/- 3

= 1/2 , 6 1/2 = e^x

Answer 4

Let w=e^x

w^2 - 7 w+10=0

solve for the values of w by the quadratic formula or factoring

(w-2)(w-5)

w= 2 and 5

substitute e^x for w

so

e^x = 2

and

e^x = 5

take the ln of both sides

x = ln(2)

and

x= ln(5)

are the solutions to the equation.