Solution.
( e 2 x + 1 ) + 3 − 3 ( e x + 1 ) = e x ; ( e 2 x + 1 ) + 3 − 3 e x − 3 = e x ; e 2 x − 4 e x + 1 = 0 ; (e^{2x}+1)+3-3(e^x+1)=e^x; \newline
(e^{2x}+1) +3-3e^x-3=e^x; \newline
e^{2x}-4e^x+1=0; \newline ( e 2 x + 1 ) + 3 − 3 ( e x + 1 ) = e x ; ( e 2 x + 1 ) + 3 − 3 e x − 3 = e x ; e 2 x − 4 e x + 1 = 0 ;
Lets y = e x ; y=e^x; \newline y = e x ;
y 2 − 4 y + 1 = 0 ; D = b 2 − 4 a c ; D = 16 − 4 ∗ 1 ∗ 1 = 12 ; D = 12 = 2 3 ; y 1 = ( − b + D ) / 2 a ; y 1 = ( 4 + 2 3 ) / 2 = 2 + 3 = 2 + 1.73 = 3.73 ; y 2 = ( − b − D ) / 2 a ; y 2 = ( 4 − 2 3 ) / 2 = 2 − 3 = 2 − 1.73 = 0.27 ; y^2-4y+1=0; \newline
D=b^2-4ac; \newline
D=16-4*1*1=12; \newline
\sqrt{D}=\sqrt{12}=2\sqrt{3}; \newline
y_1=(-b+\sqrt{D})/2a; \newline
y_1=(4+2\sqrt{3})/2=2+\sqrt{3}=2+1.73=3.73; \newline
y_2=(-b-\sqrt{D})/2a; \newline
y_2=(4-2\sqrt{3})/2=2-\sqrt{3}=2-1.73=0.27; \newline y 2 − 4 y + 1 = 0 ; D = b 2 − 4 a c ; D = 16 − 4 ∗ 1 ∗ 1 = 12 ; D = 12 = 2 3 ; y 1 = ( − b + D ) /2 a ; y 1 = ( 4 + 2 3 ) /2 = 2 + 3 = 2 + 1.73 = 3.73 ; y 2 = ( − b − D ) /2 a ; y 2 = ( 4 − 2 3 ) /2 = 2 − 3 = 2 − 1.73 = 0.27 ;
The main natural logarithm formula is:
l n ( e x ) = x . y 1 = e x 1 ; e x 1 = 3.73 ; l n ( e x 1 ) = l n ( 3.73 ) ; x 1 ∗ l n ( e ) = l n ( 3.73 ) ; x 1 ∗ 1 = l n ( 3.73 ) ; x 1 = 1.32 ; y 2 = e x 2 ; e x 2 = 0.27 ; l n ( e x 2 ) = l n ( 0.27 ) ; x 2 = l n ( 0.27 ) ; x 2 = − 1.3. ln(e^x)=x. \newline
y_1=e^{x_1};\newline
e^{x_1}=3.73;\newline
ln(e^{x_1})=ln(3.73);\newline
x_1*ln(e)=ln(3.73);\newline
x_1*1=ln(3.73);\newline
x_1=1.32; \newline
y_2=e^{x_2}; \newline
e^{x_2}=0.27; \newline
ln(e^{x_2})=ln(0.27); \newline
x_2=ln(0.27); \newline
x_2=-1.3. l n ( e x ) = x . y 1 = e x 1 ; e x 1 = 3.73 ; l n ( e x 1 ) = l n ( 3.73 ) ; x 1 ∗ l n ( e ) = l n ( 3.73 ) ; x 1 ∗ 1 = l n ( 3.73 ) ; x 1 = 1.32 ; y 2 = e x 2 ; e x 2 = 0.27 ; l n ( e x 2 ) = l n ( 0.27 ) ; x 2 = l n ( 0.27 ) ; x 2 = − 1.3.
Answer: x 1 = 1.32 ; x 2 = − 1.3. x_1=1.32; x_2=-1.3. x 1 = 1.32 ; x 2 = − 1.3.
#339914 on Dec 2023
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