Soit
x x x un réel strictement positif.
ln ( x 2 ) − ln ( x 5 e ) + ln ( 2 ) = ln ( 2 x ) + 5 \ln \left(x^{2} \right)-\ln \left(\frac{x^{5} }{e} \right)+\ln \left(2\right)=\ln \left(2x\right)+5 ln ( x 2 ) − ln ( e x 5 ) + ln ( 2 ) = ln ( 2 x ) + 5 . Nous commençons par mettre tous les logarithmes à gauche du signe égal.
ln ( x 2 ) − ln ( x 5 e ) + ln ( 2 ) − ln ( 2 x ) = 5 {\color{blue}\ln \left(x^{2} \right)-\ln \left(\frac{x^{5} }{e} \right)}+\ln \left(2\right)-\ln \left(2x\right)=5 l n ( x 2 ) − l n ( e x 5 ) + ln ( 2 ) − ln ( 2 x ) = 5 ln ( a ) − ln ( b ) = ln ( a b ) \ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right) ln ( a ) − ln ( b ) = ln ( b a ) ln ( x 2 ( x 5 e ) ) + ln ( 2 ) − ln ( 2 x ) = 5 {\color{blue}\ln \left(\frac{x^{2} }{\left(\frac{x^{5} }{e} \right)} \right)}+\ln \left(2\right)-\ln \left(2x\right)=5 l n ⎝ ⎛ ( e x 5 ) x 2 ⎠ ⎞ + ln ( 2 ) − ln ( 2 x ) = 5 ln ( x 2 × e x 5 ) + ln ( 2 ) − ln ( 2 x ) = 5 \ln \left(x^{2} \times \frac{e}{x^{5} } \right)+\ln \left(2\right)-\ln \left(2x\right)=5 ln ( x 2 × x 5 e ) + ln ( 2 ) − ln ( 2 x ) = 5 ln ( e x 3 ) + ln ( 2 ) − ln ( 2 x ) = 5 \ln \left(\frac{e}{x^{3} } \right)+{\color{blue}\ln \left(2\right)-\ln \left(2x\right)}=5 ln ( x 3 e ) + l n ( 2 ) − l n ( 2 x ) = 5 ln ( a ) − ln ( b ) = ln ( a b ) \ln \left(a\right)-\ln \left(b\right)=\ln \left(\frac{a}{b} \right) ln ( a ) − ln ( b ) = ln ( b a ) ln ( e x 3 ) + ln ( 2 2 x ) = 5 \ln \left(\frac{e}{x^{3} } \right)+{\color{blue}\ln \left(\frac{2}{2x} \right)}=5 ln ( x 3 e ) + l n ( 2 x 2 ) = 5 ln ( e x 3 ) + ln ( 1 x ) = 5 \ln \left(\frac{e}{x^{3} } \right)+\ln \left(\frac{1}{x} \right)=5 ln ( x 3 e ) + ln ( x 1 ) = 5 ln ( a ) + ln ( b ) = ln ( a × b ) \ln \left(a\right)+\ln \left(b\right)=\ln \left(a\times b \right) ln ( a ) + ln ( b ) = ln ( a × b ) ln ( e x 3 × 1 x ) = 5 \ln \left(\frac{e}{x^{3} } \times \frac{1}{x} \right)=5 ln ( x 3 e × x 1 ) = 5 ln ( e x 4 ) = 5 \ln \left(\frac{e}{x^{4} } \right)=5 ln ( x 4 e ) = 5 ln e a = a \ln e^{a }=a ln e a = a ln ( e x 4 ) = ln ( e 5 ) \ln \left(\frac{e}{x^{4} } \right)=\ln \left(e^{5} \right) ln ( x 4 e ) = ln ( e 5 ) ln ( A ) = ln ( B ) ⇔ A = B \ln \left(A\right)=\ln \left(B\right)\Leftrightarrow A=B ln ( A ) = ln ( B ) ⇔ A = B e x 4 = e 5 \frac{e}{x^{4} } =e^{5} x 4 e = e 5 e x 4 = e 5 1 \frac{e}{x^{4} } =\frac{e^{5} }{1} x 4 e = 1 e 5 a b = c d ⇔ b a = d c \frac{a}{b} =\frac{c}{d} \Leftrightarrow \frac{b}{a} =\frac{d}{c} b a = d c ⇔ a b = c d x 4 e = 1 e 5 \frac{x^{4} }{e} =\frac{1}{e^{5} } e x 4 = e 5 1 x 4 = 1 e 5 × e x^{4} =\frac{1}{e^{5} } \times e x 4 = e 5 1 × e x 4 = e e 5 x^{4} =\frac{e}{e^{5} } x 4 = e 5 e x 4 = e e 4 × e x^{4} =\frac{e}{e^{4} \times e} x 4 = e 4 × e e car
e 5 = e 4 × e e^{5}=e^{4} \times e e 5 = e 4 × e x 4 = 1 e 4 x^{4} =\frac{1}{e^{4} } x 4 = e 4 1 x 4 = ( 1 e ) 4 x^{4} =\left(\frac{1}{e} \right)^{4} x 4 = ( e 1 ) 4 X 4 = A 4 ⇔ X = A X^{4} =A^{4} \Leftrightarrow X=A X 4 = A 4 ⇔ X = A quand
X X X est strictement positif .