17^{x-1}=e^{2x+1}

Lösung

1 7^{x - 1} = e^{2 x + 1}

x = \frac{1 + ln ( 17 )}{ln ( 17 ) - 2}

Dezimale

x = 4.60051 \dots

Schritte zur Lösung

1 7^{x - 1} = e^{2 x + 1}

Wende Exponentenregel an

(x - 1) ln (17) = 2 x + 1

Löse

(x - 1) ln (17) = 2 x + 1 : x = \frac{1 + ln ( 17 )}{ln ( 17 ) - 2}

x = \frac{1 + ln ( 17 )}{ln ( 17 ) - 2}

((1 6^{x})^{+})^{2} = \frac{1}{4}

2^{x + 1} - 2^{3 - x} = 6

2^{\frac{1}{2}} + 2^{\frac{1}{3}} = 2^{x}

1.0 8^{x} = 2

(\frac{1}{2})^{x} = 0.26