Solution 3.4:2b

From Förberedande kurs i matematik 1

If we write the equation as

ex2+ex=4

we see that x appears only in the combination ex and it is therefore appropriate to treat ex as a new unknown in the equation and then, when we have obtained the value of ex, we can calculate the corresponding value of x by simply taking the logarithm.

For clarity, we set t=ex, so that the equation can be written as

and we solve this second-degree equation by completing the square,

t2+t=t+212−212=t+212−41

which gives

t+212−41=4t=−21217.

These two roots give us two possible values for ex,

ex=−21−217orex=−21+217.

In the first case, the right-hand side is negative and because "e raised to anything" can never be negative, there is no x that can satisfy this equality. The other case, on the other hand, has a positive right-hand side (because 171 ) and we can take the logarithm of both sides to obtain

x=ln217−21.

Note: It is a little tricky to check the answer to the original equation, so we can be satisfied with substituting t=172−12  into the equation t2+t=4,

LHS=217−212+217−21=417−221217+41+217−21=417+41−21=417+1−2=416=4=RHS.