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 = e⁴

x = e⁴-30

 

b)

log(x²+x+15) = log(2x²+2x+3)

x²+x+15 = 2x²+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