5.) $$ 16^{x}-1 $$ $$ 5 imes 4^{x}-5 $$

Question

5.) $$ 16^{x}-1 $$ $$ 5 	imes 4^{x}-5 $$

UpStudy AI · Accepted Answer

To solve the equation $$ 16^{x} - 1 = 5 \times 4^{x} - 5 $$, we will first rewrite the terms in a more manageable form.

1. **Rewrite the bases**: Notice that $$ 16 $$ can be expressed as $$ 4^2 $$. Therefore, we can rewrite $$ 16^x $$ as $$ (4^2)^x = 4^{2x} $$.

2. **Substitute**: The equation now becomes:
   $$
   4^{2x} - 1 = 5 \times 4^{x} - 5
   $$

3. **Rearrange the equation**: Move all terms to one side:
   $$
   4^{2x} - 5 \times 4^{x} + 4 = 0
   $$

4. **Let $$ y = 4^{x} $$**: This substitution simplifies our equation to:
   $$
   y^2 - 5y + 4 = 0
   $$

5. **Solve the quadratic equation**: We can use the quadratic formula $$ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$ a = 1, b = -5, c = 4 $$.

Now, let's calculate the discriminant and the roots.

Calculating the discriminant:
$$
b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot 4 = 25 - 16 = 9
$$

Now, applying the quadratic formula:
$$
y = \frac{5 \pm \sqrt{9}}{2 \cdot 1} = \frac{5 \pm 3}{2}
$$

Calculating the two possible values for $$ y $$:
1. $$ y = \frac{5 + 3}{2} = \frac{8}{2} = 4 $$
2. $$ y = \frac{5 - 3}{2} = \frac{2}{2} = 1 $$

6. **Back substitute for $$ x $$**:
   - For $$ y = 4 $$:
     $$
     4^{x} = 4 \implies x = 1
     $$
   - For $$ y = 1 $$:
     $$
     4^{x} = 1 \implies x = 0
     $$

Thus, the solutions to the equation $$ 16^{x} - 1 = 5 \times 4^{x} - 5 $$ are:
$$
\boxed{0 \text{ and } 1}
$$
The solutions are $$ x = 0 $$ and $$ x = 1 $$.

5.) \( 16^{x}-1 \) \( 5 \times 4^{x}-5 \)

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