Certainly! To solve the equation [tex]\(16x + 8 = 24 \left(4^{x-1}\right)\)[/tex], follow these steps:
1. Identify and simplify the components:
The given equation is:
[tex]\[
16x + 8 = 24 \left( 4^{x-1} \right)
\][/tex]
2. Rewrite the exponential term using properties of exponents:
Note that [tex]\(4 = 2^2\)[/tex], so:
[tex]\[
4^{x-1} = (2^2)^{x-1} = 2^{2(x-1)} = 2^{2x-2}
\][/tex]
Therefore, the equation becomes:
[tex]\[
16x + 8 = 24 \cdot 2^{2x-2}
\][/tex]
3. Attempt to isolate the exponential term:
Rewrite the exponential term for easier handling:
[tex]\[
24 \cdot 2^{2x-2} = 24 \cdot \frac{2^{2x}}{2^2} = 6 \cdot 2^{2x}
\][/tex]
Now, the equation is:
[tex]\[
16x + 8 = 6 \cdot 2^{2x}
\][/tex]
4. Explore possible simple solutions directly:
- For [tex]\(x = 1\)[/tex]:
[tex]\[
16(1) + 8 = 24 \cdot 4^{1-1}
\][/tex]
[tex]\[
16 + 8 = 24 \cdot 1
\][/tex]
[tex]\[
24 = 24
\][/tex]
Thus, [tex]\(x = 1\)[/tex] is a solution.
5. Consider the transformation to logarithmic form for a more general solution:
Rewrite the equation separating the components not involving [tex]\(x\)[/tex]:
[tex]\[
16x + 8 = 6 \cdot 2^{2x} \implies 6 \cdot 2^{2x} - 16x - 8 = 0
\][/tex]
Here, we notice a sophisticated transcendental equation form that general solutions might involve special functions, particularly Lambert W function.
6. Apply Lambert W function (if familiar) or recognize complex transcendental solutions:
Suppose we solve the equation [tex]\(6 \cdot 2^{2x} = 16x + 8\)[/tex]:
By recognizing that exact algebraic solutions are complex, solutions involve using Lambert W function to express formally:
[tex]\[
2^{2x} = \frac{16x + 8}{6} \implies e^{2x \ln 2} = \frac{16x+8}{6}
\][/tex]
Transform for standard form involving Lambert W:
Recognize complexity and finalize;
Therefore the general solution involving Lambert W function can be:
[tex]\[
x = \frac{-\log(2) - LambertW(-3\ln(2)/8)}{2\ln(2)} \quad (alongside ~ x = 1)
\][/tex]
The complete solutions to the equation are:
[tex]\[
x = 1, \quad \text{and} \quad x = \frac{-\log(2) - LambertW\left(-\frac{3 \log(2)}{8}\right)}{2\log(2)}
\][/tex]
