$$ \left. \begin{array} { l l } { 1.2 .2 } & { 3 ^ { x } - 3 ^ { 2 - x } = 8 } \ { 1.2 .3 } & { x - 4 x ^ { \frac { 1 } { 2 } } - 5 = 0 } \end{array} ight. $$

Question

$$ \left. \begin{array} { l l } { 1.2 .2 } & { 3 ^ { x } - 3 ^ { 2 - x } = 8 } \ { 1.2 .3 } & { x - 4 x ^ { \frac { 1 } { 2 } } - 5 = 0 } \end{array} ight. $$

UpStudy AI · Accepted Answer

**For Equation $$ 3^x - 3^{2-x} = 8 $$:**

1. Rewrite the term $$ 3^{2-x} $$ as:
   $$
   3^{2-x} = \frac{3^2}{3^x} = \frac{9}{3^x}.
   $$
2. Let $$ y = 3^x $$. Then the equation becomes:
   $$
   y - \frac{9}{y} = 8.
   $$
3. Multiply both sides by $$ y $$ (noting that $$ y > 0 $$):
   $$
   y^2 - 9 = 8y.
   $$
4. Rearrange to obtain a quadratic equation:
   $$
   y^2 - 8y - 9 = 0.
   $$
5. Solve the quadratic using the quadratic formula:
   $$
   y = \frac{8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot (-9)}}{2} = \frac{8 \pm \sqrt{64+36}}{2} = \frac{8 \pm \sqrt{100}}{2}.
   $$
   $$
   y = \frac{8 \pm 10}{2}.
   $$
6. The two potential solutions are:
   $$
   y = \frac{8+10}{2} = 9 \quad \text{or} \quad y = \frac{8-10}{2} = -1.
   $$
   Since $$ y = 3^x > 0 $$, we discard $$ y = -1 $$.
7. Thus, we have:
   $$
   3^x = 9 = 3^2,
   $$
   which implies:
   $$
   x = 2.
   $$

---

**For Equation $$ x - 4\sqrt{x} - 5 = 0 $$:**

1. Let $$ t = \sqrt{x} $$. Then $$ x = t^2 $$, and the equation becomes:
   $$
   t^2 - 4t - 5 = 0.
   $$
2. Solve the quadratic equation using the quadratic formula:
   $$
   t = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-5)}}{2} = \frac{4 \pm \sqrt{16+20}}{2} = \frac{4 \pm \sqrt{36}}{2}.
   $$
   $$
   t = \frac{4 \pm 6}{2}.
   $$
3. The potential solutions for $$ t $$ are:
   $$
   t = \frac{4+6}{2} = 5 \quad \text{or} \quad t = \frac{4-6}{2} = -1.
   $$
   Since $$ t = \sqrt{x} $$ must be nonnegative, we discard $$ t = -1 $$.
4. Therefore, we have:
   $$
   \sqrt{x} = 5.
   $$
5. Squaring both sides gives:
   $$
   x = 25.
   $$

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**Final Answers:**

- For $$ 3^x - 3^{2-x} = 8 $$, the solution is $$ x = 2 $$.
- For $$ x - 4\sqrt{x} - 5 = 0 $$, the solution is $$ x = 25 $$.
The solutions are: - $$ x = 2 $$ for the first equation. - $$ x = 25 $$ for the second equation.

\( \left. \begin{array} { l l } { 1.2 .2 } & { 3 ^ { x } - 3 ^ { 2 - x } = 8 } \\ { 1.2 .3 } & { x - 4 x ^ { \frac { 1 } { 2 } } - 5 = 0 } \end{array} \right. \)

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