Question 816164
It is "ell-en" not "eye-en":
{{{e^ln(x+1)^2=x+3e^0}}}
And I am guessing that the exponent of 2 applies to (x+1) not to ln(x+1):
{{{e^ln((x+1)^2)=x+3e^0}}}
not:
{{{e^((ln(x+1)))^2=x+3e^0}}}
If my guess is wrong then you'll have to re-post. (If so, then use parentheses to make it clear.<br>
{{{e^ln((x+1)^2)=x+3e^0}}}
The left side simplifies easily if you understand what logarithms are. Logarithms are exponents. {{{log(4, (16))}}} is the exponent one would put on a 4 to get 16 as the result. In general, {{{log(a, (p))}}} 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 {{{ln((x+1)^2)}}} represents the exponent one would put on an "e" to get {{{(x+1)^2}}} as a result.<br>
And look at where the {{{ln((x+1)^2)}}} is in our equation. It is the exponent on an "e"!! So {{{e^ln((x+1)^2)}}} is simply {{{(x+1)^2}}}!! This makes our equation:
{{{(x+1)^2=x+3e^0}}}
And since any number (except zero) to the zero power is 1, the equation simplifies further:
{{{(x+1)^2=x+3}}}<br>
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 {{{(a+b)^2 = a^2+2ab+b^2}}} pattern:
{{{(x)^2+2(x)(1)+ (1)^2=x+3}}}
which simplifies to:
{{{x^2+2x+ 1=x+3}}}
Now we want a zero on one side. Subtracting x and 3 from each side:
{{{x^2+x-2=0}}}
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<br>
Last we check. Use the original equation to check:
{{{e^ln((x+1)^2)=x+3e^0}}}
Checking x = -2:
{{{e^ln((-2)+1)^2)=(-2)+3e^0}}}
Simplifying...
{{{e^ln((-1)^2)=(-2)+3e^0}}}
{{{e^ln(1)=(-2)+3e^0}}}
Since the exponent for "e" that results in 1 is zero, ln(1) = 0:
{{{e^0=(-2)+3e^0}}}
The e to the zero powers are 1's:
{{{1=(-2)+3}}}
{{{1=1}}} Check!!
Checking x = 1:
{{{e^ln((1)+1)^2)=(1)+3e^0}}}
Simplifying...
{{{e^ln(2^2)=(1)+3e^0}}}
{{{e^ln(4)=(-2)+3e^0}}}
Since ln(4) is the exponent for "e" that results in 4 and since it is the exponent on an "e":
{{{4=(1)+3e^0}}}
The e to the zero power is 1:
{{{4=(1)+3}}}
{{{4=4}}} Check!!<br>
P.S. It might be useful to memorize that {{{a^log(a, (x)) = x}}} for all bases (including "e").