Pangkat Rasional Soal Penggemar Matematika

Soal Penggemar Matematika

1. Tentukan nilai x yang memenuhi persamaan berikut.
   1. 3^{5x - 1}= 27^{x + 3}
   2. ${\left(\frac{1}{2}\right)}^{\mathrm{2x\; +\; 1}}=\sqrt{\frac{2^{4x\; -\; 1}}{128}}$
   3. $\sqrt{3^{2x\; +\; 1}}={\left(27\right)}^{\frac{1}{3}}$
   4. ${\left(\frac{1}{3}\right)}^2\sqrt{3^{2x\; +\; 1}}=27$
   5. $\sqrt{3^{2x\; +\; 1}}=9^{x\; -\; 2}$
   6. ${\left(\frac{1}{3}\right)}^{2x\; -\; 1}=\frac{1}{27}$
   7. $\frac{3^{2x}}{3}=27$

Jawab :

1. 1. 3^{5x - 1} = 27^{x + 3}
      3^{5x - 1} = 3^{3(x + 3)}
      3^{5x - 1} = 3^{3x + 9}
      5x - 1 = 3x + 9
      5x - 3x = 9 + 1
      2x = 10
      x = 5
      
   
   
   2. ${\left(\frac{1}{2}\right)}^{\mathrm{2x\; +\; 1}}=\sqrt{\frac{2^{4x\; -\; 1}}{128}}$
      $2^{\mathrm{-1(2x\; +\; 1)}}={\left(\frac{2^{4x\; -\; 1}}{2^7}\right)}^{\frac{1}{2}}$
      $2^{\mathrm{-2x\; -\; 1}}={\left(2^{4x\; -\; 1\; -\; 7}\right)}^{\frac{1}{2}}$
      $2^{\mathrm{-2x\; -\; 1}}={\left(2^{4x\; -\; 8}\right)}^{\frac{1}{2}}$
      $2^{\mathrm{-2x\; -\; 1}}=2^{(4x\; -\; 8)\frac{1}{2}}$
      $2^{\mathrm{-2x\; -\; 1}}=2^{2x\; -\; 4}$
      -2x - 1 = 2x - 4
      -2x - 2x = -4 + 1
      -4x = 3
      x = $-\frac{3}{4}$
   
   
   3. $\sqrt{3^{2x\; +\; 1}}={\left(27\right)}^{\frac{1}{3}}$
      $3^{\frac{2x\; +\; 1}{2}}={\left(3^3\right)}^{\frac{1}{3}}$
      $3^{\frac{2x\; +\; 1}{2}}=3^1$
      $\frac{2x\; +\; 1}{2}=1$
      2x + 1 = 2
      2x = 2 - 1
      2x = 1
      $x=\frac{1}{2}$
   
   
   4. ${\left(\frac{1}{3}\right)}^2\sqrt{3^{2x\; +\; 1}}=27$
      $3^{(-1)2}{3}^{\frac{2x\; +\; 1}{2}}=3^3$
      $3^{-2+\frac{2x\; +\; 1}{2}}=3^3$
      $-2+\frac{2x\; +\; 1}{2}=3$
      $\frac{2x\; +\; 1}{2}=3+2$
      $\frac{2x\; +\; 1}{2}=5$
      2x + 1 = 10
      2x = 10 - 1
      2x = 9
      $x=\frac{9}{2}$
      $x=4\frac{1}{2}$
      
   
   
   5. $\sqrt{3^{2x\; +\; 1}}=9^{x\; -\; 2}$
      $3^{\frac{2x\; +\; 1}{2}}=3^{2(x\; -\; 2)}$
      $3^{\frac{2x\; +\; 1}{2}}=3^{2x\; -\; 4}$
      $\frac{2x\; +\; 1}{2}=2x\; -\; 4$
      2x + 1 = 2(2x - 4)
      2x + 1 = 4x - 8
      2x - 4x = -8 - 1
      -2x = -9
      $x=\frac{-9}{-2}$
      $x=4\frac{1}{2}$
      
   
   
   6. ${\left(\frac{1}{3}\right)}^{2x\; -\; 1}=\frac{1}{27}$
      $3^{-1(2x\; -\; 1)}=\frac{1}{3^3}$
      $3^{-2x\; +\; 1}=3^{-3}$
      -2x + 1 = -3
      -2x = -3 - 1
      -2x = -4
      x = 2
   
   
   7. $\frac{3^{2x}}{3}=27$
      $3^{2x\; -\; 1}=3^3$
      2x - 1 = 3
      2x = 3 + 1
      2x = 4
      x = 2