The equation `2^(x +2) +4^x + 4 = 0` has to be solved for x.
Use the following property of exponents: `a^x*a^y = a^(x+y)` , `(a^x)^y = (a^y)^x`
`2^(x +2) +4^x + 4 = 0`
=> `2^x*2^2 + (2^2)^x + 4 = 0`
=> `4*2^x + (2^x)^2 + 2^2 = 0`
Let `2^x = y`
=> `4y + y^2 + 2^2 = 0`
=> `y^2 + 2*2*y + 2^2 = 0`
=> `(y + 2)^2 = 0`
=> `y = -2`
But `y = 2^x` and as the base 2 is positive for no value of x is `2^x` negative.
As `2^x != -2` , the equation `2^(x +2) +4^x + 4 = 0` does not have a solution.
You have to determine if `2^(x +2) +4^x + 4 = 0` has a solution.
Use the rules: `x^(a+b) = x^a*x^b` and `x^a = (x^(a/2))^(1/2)`
`2^(x +2) +4^x + 4 = 0` can be written as
`2^x*2^2 + 4^x + 4 = 0`
`2^x*4 + (2^2)^x + 4 = 0`
`2^x*4 + (2^x)^2 + 4 = 0`
`(2^x)^2 + 4*2^x + 4 = 0`
This is of the form `(a + 2) = a^2 + 4a + 4`
`(2^x + 2)^2 = 0`
`2^x + 2 = 0`
`2^x = -2`
But it is not possible to raise a positive number to any power and obtain a negative result.
Therefore the equation `2^(x +2) +4^x + 4 = 0` does not have a solution.
This problem tests the concept of factoring with exponents.
2^(x+2) + 4^(x) + 4 = 0
Since for exponents, x^(y+z) can be rewritten as (x^y)*(x^z) and vice versa, the first term of the problem can be rewritten to make the equation as follows.
(2^x)*(2^2) + 4^(x) + 4 = 0
Additionally, since 4 can be rewritten as a base of 2, the second term can be changed to make the equation look like this.
(2^x)*(2^2) + (2^2)^x + 4 = 0
Since (x^y)^z is equivalent to (x^z)^y, the second term can be rewritten once again.
(2^x)*(2^2) + (2^x)^2 + 4 = 0
Now, this is a simple quadratic equation in disguise. If you replace (2^x) with simply "x" and multiply through the (2^2) in the first term, it can be rewritten as
4x + x^2 + 4 = 0 --> x^2 + 4x + 4 = 0
This can be factored to
(x+2)(x+2) = 0
Replacing the "x" with (2^x) yields
((2^x)+2) = 0
Because of the zero power property,
(2^x) + 2 = 0
(2^x) = -2
Since a positive number raised to any exponent can never equal a negative number, there is no solution to this problem.
For this problem I would use either the property of logs or natural logs, both will get you the same answer.
The third law/or power rule of both states that if you take the log/ln or a number/variable with a power to a # or variable then that power can be moved to the front of the log, leaving you with the following. (btw, I moved the last 4 over b/f I took the logs/ln)
(x+2)log2 + xlog4 = -log4
xlog2 +2log2 +xlog4 =-log4
x(log2 + log4) = -(log4 + 2log2)
x = -(log4 + 2log2) / (log2 + log4)
You can now use the first two rules that state log addition is equivalent to multiplication if the bases are the same, and that subtraction is equivalent to division given the same requirement.
x = -(log4 +(log2 +log2))/(log2 +log4)
x = -(log4 + log 4)/ (log2 + log4)
x = - log16/log8
plug into calculator