7^{5x^2-42x}=10^{125-42x}

Lösung

7^{5 x^{2} - 42 x} = 1 0^{125 - 42 x}

x = \frac{42 ln ( 7 ) - 42 ln ( 10 ) + 1764 ln ^{2} ( 10 ) - 1028 ln ( 7 ) ln ( 10 ) + 1764 ln ^{2} ( 7 )}{10 ln ( 7 )}, x = \frac{42 ln ( 7 ) - 42 ln ( 10 ) - 1764 ln ^{2} ( 10 ) - 1028 ln ( 7 ) ln ( 10 ) + 1764 ln ^{2} ( 7 )}{10 ln ( 7 )}

Dezimale

x = 4.72334 \dots, x = - 6.26301 \dots

Schritte zur Lösung

7^{5 x^{2} - 42 x} = 1 0^{125 - 42 x}

Wende Exponentenregel an

(5 x^{2} - 42 x) ln (7) = (125 - 42 x) ln (10)

Löse

(5 x^{2} - 42 x) ln (7) = (125 - 42 x) ln (10) : x = \frac{42 ln ( 7 ) - 42 ln ( 10 ) + 1764 ln ^{2} ( 10 ) - 1028 ln ( 7 ) ln ( 10 ) + 1764 ln ^{2} ( 7 )}{10 ln ( 7 )}, x = \frac{42 ln ( 7 ) - 42 ln ( 10 ) - 1764 ln ^{2} ( 10 ) - 1028 ln ( 7 ) ln ( 10 ) + 1764 ln ^{2} ( 7 )}{10 ln ( 7 )}

x = \frac{42 ln ( 7 ) - 42 ln ( 10 ) + 1764 ln ^{2} ( 10 ) - 1028 ln ( 7 ) ln ( 10 ) + 1764 ln ^{2} ( 7 )}{10 ln ( 7 )}, x = \frac{42 ln ( 7 ) - 42 ln ( 10 ) - 1764 ln ^{2} ( 10 ) - 1028 ln ( 7 ) ln ( 10 ) + 1764 ln ^{2} ( 7 )}{10 ln ( 7 )}

so l v e f or t, e^{k t} = \frac{1}{7}

3 ∣ 2^{x} - 1 ∣ = 2^{x}

3 x e^{x} + x^{2} e^{x} = 0

e^{4 x} = 2

2^{5 x + 3} = 3^{2 x + 1}