SOLUTION: Solve and check e^In(x+1)^2=x+3e^0

Question 816164: Solve and check
e^In(x+1)^2=x+3e^0
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
It is "ell-en" not "eye-en":

And I am guessing that the exponent of 2 applies to (x+1) not to ln(x+1):

not:

If my guess is wrong then you'll have to re-post. (If so, then use parentheses to make it clear.

The left side simplifies easily if you understand what logarithms are. Logarithms are exponents. is the exponent one would put on a 4 to get 16 as the result. In general, is the exponent one would put on an "a" to get "p" as a result. ln's are just logarithms with a special base, e. So ln(7) is the exponent one would put on an "e" to get 7 as the result. And represents the exponent one would put on an "e" to get as a result.

And look at where the is in our equation. It is the exponent on an "e"!! So is simply !! This makes our equation:

And since any number (except zero) to the zero power is 1, the equation simplifies further:

We now have a quadratic equation to solve. First we square the left side. We could use FOIL for this but I prefer to use the pattern:

which simplifies to:

Now we want a zero on one side. Subtracting x and 3 from each side:

Now we factor:
(x+2)(x-1) = 0
Next the Zero Product Property:
x+2 = 0 or x-1 = 0
Solving these we get:
x = -2 or x = 1

Last we check. Use the original equation to check:

Checking x = -2:

Simplifying...

Since the exponent for "e" that results in 1 is zero, ln(1) = 0:

The e to the zero powers are 1's:

Check!!
Checking x = 1:

Simplifying...

Since ln(4) is the exponent for "e" that results in 4 and since it is the exponent on an "e":

The e to the zero power is 1:

Check!!

P.S. It might be useful to memorize that for all bases (including "e").
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