2^n+3^m=a^k

Lösung

2^{n} + 3^{m} = a^{k}

k = \frac{ln ( 2 ^{n} + 3 ^{m} )}{ln ( a )}

Schritte zur Lösung

2^{n} + 3^{m} = a^{k}

Wende Exponentenregel an

ln (2^{n} + 3^{m}) = k ln (a)

Löse

ln (2^{n} + 3^{m}) = k ln (a) : k = \frac{ln ( 2 ^{n} + 3 ^{m} )}{ln ( a )}

k = \frac{ln ( 2 ^{n} + 3 ^{m} )}{ln ( a )}

3^{x} = 8

(2^{x}) (3^{x - 2}) = 24

(3^{2 x + 1}) \cdot (2^{x + 1}) = 6

so l v e f or m, 144 = \frac{( 2 ^{m} )}{( 3 ^{n} )}

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