log_{2}(2x+1)=4

Lösung

lo g_{2} (2 x + 1) = 4

x = \frac{15}{2}

Dezimale

x = 7.5

Schritte zur Lösung

lo g_{2} (2 x + 1) = 4

Wende die log Regel an

2 x + 1 = 16

Löse

2 x + 1 = 16 : x = \frac{15}{2}

x = \frac{15}{2}

so l v e f or x, 0 = ln (1 - y^{2} - x)

3.9 = ln (\frac{( k - x )}{x})

ln (4) + 2 ln (5) = ln (x)

0 = lo g_{3} (x + 1) - 2

ln (x) = m - 1