(7) $$ 0,25^{2-x}=\frac{32}{2^{x+3}} $$ Solve for $$ x $$ : (1) $$ 2^{x}+2^{x+1}=24 $$ (3) $$ 3^{x+1}-2 \cdot 3^{x-1}=21 $$ (5) $$ 3^{2 x}-4 \cdot 3^{x}+3=0 $$

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Solve the equation $$ 0.25^{2-x}=\frac{32}{2^{x+3}} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$0.25^{2-x}=\frac{32}{2^{x+3}}$$
- step1: Convert the expressions:
  $$\left(\frac{1}{4}\right)^{2-x}=2^{2-x}$$
- step2: Rewrite the expression:
  $$2^{-4+2x}=2^{2-x}$$
- step3: Set the exponents equal:
  $$-4+2x=2-x$$
- step4: Move the expression to the left side:
  $$2x+x=2+4$$
- step5: Add and subtract:
  $$3x=2+4$$
- step6: Add and subtract:
  $$3x=6$$
- step7: Divide both sides:
  $$\frac{3x}{3}=\frac{6}{3}$$
- step8: Divide the numbers:
  $$x=2$$
The solution to the equation $$0.25^{2-x}=\frac{32}{2^{x+3}}$$ is $$x=2$$.
$$ x = 2 $$

(7)
Solve for :
(1)
(3)
(5)

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