4^x+1 - 9(2^x) = -2. Interprete: 4, 9(2) and -2 are the base while x+1 and x are the powers.

talk true na assignment

impossible! no root of negative number exist

Yeah, an assignment.

X=0 and X=-2

Princenovo:
4^x+1 - 9(2^x) = -2. Interprete: 4, 9(2) and -2 are the base while x+1 and x are the powers.

Is this the question?
4^x+1 - 9(2^x) = - 2
If so the answers are: x= -2; x=1.

Solution:
4^x+1 - 9(2^x = -2
The number 4 can be written as 2²
(2²)^x+1 -9(2^x) = -2
From the laws of logarithm: (a^b)^m = a^bm
So we have:
2^2×+2 -9(2^×) = -2
From the laws of logarithm a^m+n = a^m * aⁿ
So we have:
2^2x * 2² - 9(2^×/) =-2
(2^x)² * 4 -9(2^{×/sup]) = -2
Since 2[sup]x} is common, to simplify we replace it by a variable, let's say y.
2^x = y
We substitute y for 2^x
Then we have:
(y)² *4 -9(y) = -2
4y² - 9y = -2.
Solve the quadratic equation you get:
y = 2, y=1/4
Substituting back
Since y = 2^x
When y = 2
2=2^x
2¹ = 2^×
x = 1.

When y = 1/4
2^x = 1/4
2^x =4^-1
2^× = 2^-2
x = -2.

Therefore, x = 1, x = -2.
Any questions?
Next time pay attention in class.