Lösen Sie diese Exponentialgleichung 2^x - 3^x = 3^{x-2} - 2^{x-1}

Question

$$ { 2 }^{ x }-{ 3 }^{ x }\quad =\quad { 3 }^{ x-2 }-{ 2 }^{ x-1 } $$

Answer 1

2^x - 3^x = 3^{x-2} - 2^{x-1} 

2^x + 2^{x-1}  = 3^{x-2} + 3^x       | Potenzgesetze

2^x + 1/2 * 2^{x}  = 1/9 * 3^{x} + 3^x 

3/2 * 2^x = 10/9 * 3^x

3/2 * 9/10 = (3/2)^x

27/20 = (3/2)^x

x = ln(27/20) / ln(3/2)

Bitte erst nachrechnen und bei Bedarf Logarithmen vereinfachen oder Taschenrechner verwenden.

Answer 2

$$  x = \frac{ \ln(20) - \ln(27)  }{ \ln(2) - \ln(3)  } \approx  0.74  $$

Answer 3

2^x - 3^x = 3^{x - 2} - 2^{x - 1}

2^x + 2^{x - 1} = 3^x + 3^{x - 2}

2^x + 2^x / 2 = 3^x + 3^x / 9

2^x·(1 + 1/2) = 3^x·(1 + 1/9)

2^x / 3^x = (1 + 1/9) / (1 + 1/2)

(2/3)^x = 20/27

x = LN(20/27) / LN(2/3) = 0.7401