solvefor x,4^{x+y}=3^{x^2+y^2}

Lösung

löse nach

x, 4^{x + y} = 3^{x^{2} + y^{2}}

x = \frac{ln ( 2 ) + - ln ( 3 ) y ( ln ( 3 ) y - 2 ln ( 2 ) ) + ln ^{2} ( 2 )}{ln ( 3 )}, x = \frac{ln ( 2 ) - - ln ( 3 ) y ( ln ( 3 ) y - 2 ln ( 2 ) ) + ln ^{2} ( 2 )}{ln ( 3 )}

Schritte zur Lösung

4^{x + y} = 3^{x^{2} + y^{2}}

Wende Exponentenregel an

(x + y) ln (4) = (x^{2} + y^{2}) ln (3)

Löse

(x + y) ln (4) = (x^{2} + y^{2}) ln (3) : x = \frac{ln ( 2 ) + - ln ( 3 ) y ( ln ( 3 ) y - 2 ln ( 2 ) ) + ln ^{2} ( 2 )}{ln ( 3 )}, x = \frac{ln ( 2 ) - - ln ( 3 ) y ( ln ( 3 ) y - 2 ln ( 2 ) ) + ln ^{2} ( 2 )}{ln ( 3 )}

x = \frac{ln ( 2 ) + - ln ( 3 ) y ( ln ( 3 ) y - 2 ln ( 2 ) ) + ln ^{2} ( 2 )}{ln ( 3 )}, x = \frac{ln ( 2 ) - - ln ( 3 ) y ( ln ( 3 ) y - 2 ln ( 2 ) ) + ln ^{2} ( 2 )}{ln ( 3 )}

1 0^{x} = 25

2^{l o g_{2} (x^{2})} - 3 x - 4 = 0

2^{x + 2} + 2^{x - 1} = 9

lo g_{10} (9^{x}) = 1.5

1 0^{x} = 33