Logarithmus /Exponentialfunktion : Gleichungen lösen

Question

Logarithmus und Exponentialfunktion: Lösen Sie folgende Gleichungen:

(a) \( \ln (x+30)=4 \)
(b) \( \log \left(x^{2}+x+15\right)=\log \left(2 x^{2}+2 x+3\right) \)
(c) \( e^{2 x+4}=e^{x-1} \)
(d) \( \left(e^{x}\right)^{2}=7 \)
(e) \( \log (3 x-1)+\log (x+5)=0 \)
(f) \( 2 \log (x-1)=\log (3 x+1) \)

Answer 1

Hi,

a)

ln(x+30) = 4

x+30 = e^4

x = e^4-30

 

b)

log(x^2+x+15) = log(2x^2+2x+3)

x^2+x+15 = 2x^2+2x+3

x₁ = -4 und x₂ = 3

 

c)

e^{2x+4} = e^{x-1}

2x+4 = x-1

x = -5

 

d)

(e^x)^2 = 7

e^{2x} = 7

2x = ln(7)

x = ln(7)/2

 

e)

log(3x-1)+log(x+5) = 0    Logarithmengesetz log(a)+log(b) = log(ab)

log((3x-1)(x+5)) = 0

(3x-1)(x+5) = 1

x = 1/3(√(67)-7)

 

f)

2log(x-1) = log(3x+1)

log((x-1)^2) = log(3x+1)

(x-1)^2 = 3x+1

x = 5

 

Grüße