# Find a solution for the equation e^2x + 3*e^x = 10

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We have to find the solution for e^2x + 3*e^x = 10

e^2x + 3*e^x = 10

=> e^x^2 + 3*e^x = 10

let e^x = y

=> y^2 + 3*y - 10 = 0

=> y^2 + 5y - 2y - 10 = 0

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

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

=> y = 2 and y = -5

y = e^x = 2 and y = e^x = -5

e^x = 2

take the ln of both the side

x = ln 2

ln for -5 is not defined .

**The only solution for the equation is x = ln 2**

The equation `e^(2x) + 3*e^x = 10` has to be solved for x.

`e^(2x)` can be rewritten as `(e^x)^2` .

The equation now be written as:

`(e^x)^2 + 3*e^x = 10`

`(e^x)^2 + 3*e^x - 10 = 0`

`(e^x)^2 + 5*e^x - 2e^x - 10 = 0`

`(e^x)(e^x + 5) - 2(e^x + 5) = 0`

`(e^x - 2)(e^x + 5) = 0`

`e^x - 2 = 0`

`e^x = 2`

x = ln 2

`e^x + 5 = 0`

`e^x = -5`

This is not possible as e is a positive number and the power of a positive number is always positive.

The solution of the equation `e^(2x) + 3*e^x = 10` is x = ln 2

e^2x + 3e^x = 10

First let y= e^x ==> y^2 = e^2x

Now we will substitute .

==> y^2 + 3y = 10

==> y^2 + 3y -10 = 0

Let is factor.

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

==> y1= -5 ==> e^x = -5 ==> No solution.

==> y2= 2 ==> e^x = 2 ==> x = ln 2

**Then we have one solution for the equations: x = ln 2 = 0.6931 ( approx.)**