# solve for x for (e^x+2) ln (1-2x)=0i dont know how to answer the questions when it is related to e or ln.. help me...

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in this case, i have two different ways.

1st, use inverse hyperbolic

(e^2 + e^-1 / 2 ) = 1

cosh x = 1

x = cosh^-1 1

x = [ln 1 + (1^2 -1 ) ^1/2

x=o

2nd way is more easier....

use substitution...let p=e^x and you will have a quadratic equation. try it first by yourself.. best of luck.. =)

Since the provided expression is a product of two factors, this product is zero if one or the other factor is zero.

We'll cancel each factor:

e^x + 2 = 0

e^x = -2

There is no such value of x for the exponential yields a negative value, therefore e^x + 2 > 0, but never 0.

We'll cancel the next factor:

ln(1-2x) = 0

We'll take antilogarithm, raising the natural base of logarithm, e, to the 0 power:

1 - 2x = e^0

But e^0 =1:

1 - 2x = 1

-2x = 0 => x = 0

**Since the constraint of existence of logarithm is 1 - 2x > 0 => -2x > -1 => x < 1/2 is respected, therefore the equation has the solution x = 0.**