Pangkat Rasional Soal Penggemar Matematika bagian II

 Soal Penggemar Matematika

2. Tentukan nilai x yang memenuhi persamaan berikut.
   1. 2^x . 8^{x + 2} = 64 . 4^3x
   2. 8 . 2^x = 1
   3. 2^x . 2³ = 0,25
   4. (0,04)^x . 5 = 125
   5. ${\left(\frac{1}{4}\right)}^{x\; -\; 1}=\sqrt[3]{2^{3x\; -\; 1}}$
   6. $4^{x\; +\; 3}=\sqrt[4]{8^{x\; +\; 5}}$

Jawab :

2. 1. 2^x . 8^{x + 2} = 64 . 4^3x
      2^x . 2^{3(x + 2)} = 2⁶ . 2^2(3x)
      2^x . 2^{3x + 6} = 2⁶ . 2^6x
      2^{x + 3x + 6} = 2^{6 + 6x}
      2^{4x + 6} = 2^{6 + 6x}
      4x + 6 = 6 + 6x
      4x - 6x = 6 - 6
      -2x = 0
      x = 0
   
   
   2. 8 . 2^x = 1
      2³ . 2^x = 2⁰
      2^{3 + x} = 2⁰
      3 + x = 0
      x = 0 - 3
      x = -3
   
   
   3. 2^x . 2³ = 0,25
      $2^x.2^3=\frac{1}{4}$
      $2^x.2^3=\frac{1}{2^2}$
      $2^{x\; +\; 3}=2^{-2}$
      x + 3 = -2
      x = -2 - 3
      x = -5
   
   
   4. (0,04)^x . 5 = 125
      ${\left(\frac{4}{100}\right)}^x.5=5^3$
      ${\left(\frac{1}{25}\right)}^x.5=5^3$
      ${\left(\frac{1}{5^2}\right)}^x.5=5^3$
      ${\left(5^{-2}\right)}^x.5=5^3$
      5^-2x . 5 = 5³
      5^-2x . 5¹ = 5³
      5^{-2x + 1} = 5³
      -2x + 1 = 3
      -2x = 3 - 1
      -2x = 2
      x = -1
   
   
   5. ${\left(\frac{1}{4}\right)}^{x\; -\; 1}=\sqrt[3]{2^{3x\; -\; 1}}$
      ${\left(\frac{1}{2^2}\right)}^{x\; -\; 1}=2^{\frac{3x\; -\; 1}{3}}$
      ${\left(2^{-2}\right)}^{x\; -\; 1}=2^{\frac{3x\; -\; 1}{3}}$
      $2^{-2x\; +\; 2}=2^{\frac{3x\; -\; 1}{3}}$
      $-2x\; +\; 2=\frac{3x\; -\; 1}{3}$
      $(-2x\; +\; 2)3=3x\; -\; 1$
      $-6x\; +\; 6=3x\; -\; 1$
      $-6x\; -\; 3x=-1\; -\; 6$
      $-9x=-7$
      $x=\frac{7}{9}$
      
   
   
   6. $4^{x\; +\; 3}=\sqrt[4]{8^{x\; +\; 5}}$
      $2^{2(x\; +\; 3)}=\sqrt[4]{2^{3(x\; +\; 5)}}$
      $2^{2x\; +\; 6}=2^{\frac{3x\; +\; 15}{4}}$
      $2x\; +\; 6=\frac{3x\; +\; 15}{4}$
      (2x + 6)4 = 3x + 15
      8x + 24 = 3x + 15
      8x - 3x = 15 - 24
      5x = -9
      $x=\frac{-9}{5}$
      $x=-\frac{9}{5}$
      
   
   
   7. 
      $3^{\frac{2(x\; +\; 1)}{2}}=3^{-1(x-\frac{1}{2})}$
      $3^{x\; +\; 1}=3^{-x+\frac{1}{2}}$
      $x\; +\; 1=-x+\frac{1}{2}$
      $x\; +\; x=\frac{1}{2}-1$
      $2x=-\frac{1}{2}$
      $x=-\frac{1}{4}$
      

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