स् (Solve): $$ 2^{x-2}+2^{3-x}=3 $$

Question

UpStudy AI · Accepted Answer

We start with the equation  
$$
2^{x-2} + 2^{3-x} = 3.
$$

**Step 1.** Rewrite each term by expressing the exponents in terms of $$ 2^x $$:

$$
2^{x-2} = \frac{2^x}{2^2} = \frac{2^x}{4}, \quad \text{and} \quad 2^{3-x} = \frac{2^3}{2^x} = \frac{8}{2^x}.
$$

Then the equation becomes

$$
\frac{2^x}{4} + \frac{8}{2^x} = 3.
$$

**Step 2.** Multiply both sides of the equation by 4 to eliminate the denominator:

$$
2^x + \frac{32}{2^x} = 12.
$$

**Step 3.** Introduce the substitution

$$
u = 2^x,
$$

where $$ u > 0 $$. Then the equation transforms into

$$
u + \frac{32}{u} = 12.
$$

**Step 4.** Multiply the whole equation by $$ u $$ to remove the fraction:

$$
u^2 + 32 = 12u.
$$

Rearrange the equation into standard quadratic form:

$$
u^2 - 12u + 32 = 0.
$$

**Step 5.** Solve the quadratic equation using the quadratic formula

$$
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$

where $$ a = 1 $$, $$ b = -12 $$, and $$ c = 32 $$:

$$
u = \frac{12 \pm \sqrt{144 - 128}}{2} = \frac{12 \pm \sqrt{16}}{2} = \frac{12 \pm 4}{2}.
$$

Thus, the solutions are

$$
u = \frac{12 + 4}{2} = 8 \quad \text{or} \quad u = \frac{12 - 4}{2} = 4.
$$

**Step 6.** Recall that $$ u = 2^x $$. Now solve for $$ x $$:

1. For $$ 2^x = 8 $$:
   $$
   2^x = 2^3 \quad \Longrightarrow \quad x = 3.
   $$

2. For $$ 2^x = 4 $$:
   $$
   2^x = 2^2 \quad \Longrightarrow \quad x = 2.
   $$

**Final Answer:**
$$
x = 2 \quad \text{or} \quad x = 3.
$$
$$ x = 2 $$ or $$ x = 3 $$.

स् (Solve):