Simplifier des expressions avec la fonction logarithme Exercice

A = \ln \left(3^2\right) - \ln 3 +2\ln\left(3^{\frac{1}{2}}\right)

Or \ln\left(a^n\right)=n\ln\left(a\right), on obtient donc :

A = 2\ln\left(3\right) - \ln\left(3\right) +2\times \dfrac{1}{2}\ln\left(3\right)

A = 2\ln\left(3\right) - \ln\left(3\right) +\ln\left(3\right)

A = \ln \left(\dfrac{1}{2}\right) + \ln 4

A = \ln 1 - \ln 2 + \ln 2^2

Or \ln\left(a^n\right)=nln\left(a\right), on obtient donc :

A = \ln 1 - \ln 2 +2ln2

A = \ln 7 +\ln 49

A = \ln 7 +\ln 7^2

Or \ln\left(a^n\right)=nln\left(a\right), on obtient donc :

A = \ln 7 +2ln7

A = \ln \left(\dfrac{2}{5}\right) +\ln 25

A = \ln 2 - ln5 +\ln 5^2

Or \ln\left(a^n\right)=nln\left(a\right), on obtient donc :

A = \ln 2 - ln5 +2\ln 5

A = ln2 + \ln 5

Et, comme \ln\left(a\right)+\ln\left(b\right)=\ln\left(a\times b\right) :

A = 2\ln x - \ln 3x

Or \ln\left(a^n\right)=nln\left(a\right), on obtient donc :

A = \ln x^2 - \ln 3x

A = \ln \left(\dfrac{x^2}{3x}\right)

On remarque que x^2+2x+1 =\left(x+1\right)^2

Donc A = \ln\left(x+1\right)^2 - \ln\left(x+1\right)

Or \ln\left(a^n\right)=nln\left(a\right), on obtient donc :

A = 2\ln\left(x+1\right) - \ln\left(x+1\right)

A = \ln\left(\sqrt{2x+3}\right) +\ln \left(2x+3\right)^3

A = \ln\left(\left(2x+3\right)^{\frac{1}{2}}\right) +\ln \left(\left(2x+3\right)^3\right)

Or \ln\left(a^n\right)=nln\left(a\right), on obtient donc :

A = \dfrac{1}{2}\ln\left(2x+3\right) +3\ln \left(2x+3\right)