Solving the Equation 2^2x-1-9*2^x-2+1=0
In this article, we will solve the equation 2^2x-1-9*2^x-2+1=0, which involves exponential functions and algebraic manipulations.
Step 1: Simplify the Equation
Let's start by simplifying the equation:
2^2x-1-9*2^x-2+1=0
First, we can simplify the exponentiation:
2^2x = (2^2)^x = 4^x
So, the equation becomes:
4^x - 1 - 9*2^x - 2 + 1 = 0
Step 2: Combine Like Terms
Now, let's combine like terms:
4^x - 9*2^x - 2 = 0
Step 3: Factor Out 2^x
Notice that both terms involving exponents have 2^x as a common factor. Let's factor it out:
2^x (4^x/2^x - 9) - 2 = 0
Since 4^x/2^x = 2^x, the equation becomes:
2^x (2^x - 9) - 2 = 0
Step 4: Solve for x
Now, let's solve for x. We can start by adding 2 to both sides of the equation:
2^x (2^x - 9) = 2
Next, we can divide both sides by 2:
2^x (2^x - 9) / 2 = 1
This simplifies to:
2^x - 9 = 1
Adding 9 to both sides gives:
2^x = 10
Taking the logarithm base 2 of both sides, we get:
x = log2(10)
Therefore, the solution to the equation 2^2x-1-9*2^x-2+1=0 is x = log2(10).
Conclusion
In this article, we successfully solved the equation 2^2x-1-9*2^x-2+1=0 using exponential identities and algebraic manipulations. The solution is x = log2(10).
