Sure! Let’s solve the equation [tex]\( 4^x + 2 = 6 \cdot 2^{x-1} \)[/tex] step by step.
First, recall that [tex]\( 4^x \)[/tex] can be rewritten using the properties of exponents. Since [tex]\( 4 = 2^2 \)[/tex], we have:
[tex]\[ 4^x = (2^2)^x = 2^{2x} \][/tex]
So the original equation becomes:
[tex]\[ 2^{2x} + 2 = 6 \cdot 2^{x-1} \][/tex]
Next, let’s simplify the right side of the equation:
[tex]\[ 6 \cdot 2^{x-1} = 6 \cdot \frac{2^x}{2} = 3 \cdot 2^x \][/tex]
So the equation now is:
[tex]\[ 2^{2x} + 2 = 3 \cdot 2^x \][/tex]
Let’s introduce a substitution to make the equation easier to solve. Define [tex]\( y = 2^x \)[/tex]. Thus, [tex]\( y^2 = 2^{2x} \)[/tex]. Substituting these into the equation gives:
[tex]\[ y^2 + 2 = 3y \][/tex]
Now, let’s rearrange this into a standard quadratic equation:
[tex]\[ y^2 - 3y + 2 = 0 \][/tex]
We can solve this quadratic equation using the factorization method. The quadratic factors as:
[tex]\[ (y - 1)(y - 2) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ y - 1 = 0 \quad \text{or} \quad y - 2 = 0 \][/tex]
Thus, we have two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = 1 \quad \text{or} \quad y = 2 \][/tex]
Recall that [tex]\( y = 2^x \)[/tex], so we now have two equations to solve for [tex]\( x \)[/tex]:
1. [tex]\( 2^x = 1 \)[/tex]
2. [tex]\( 2^x = 2 \)[/tex]
Solve each equation separately:
For [tex]\( 2^x = 1 \)[/tex]:
[tex]\[ 2^x = 2^0 \][/tex]
Thus,
[tex]\[ x = 0 \][/tex]
For [tex]\( 2^x = 2 \)[/tex]:
[tex]\[ 2^x = 2^1 \][/tex]
Thus,
[tex]\[ x = 1 \][/tex]
Therefore, the solutions to the equation [tex]\( 4^x + 2 = 6 \cdot 2^{x-1} \)[/tex] are:
[tex]\[ x = 0 \quad \text{and} \quad x = 1 \][/tex]
So, the solutions are [tex]\( \boxed{0 \text{ and } 1} \)[/tex].
