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...

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.