Answer to Question #199025 in Algebra for Khanya Nokilana

Question

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

Answer 1

Solution.

$(e^{2x}+1)+3-3(e^x+1)=e^x; \newline (e^{2x}+1) +3-3e^x-3=e^x; \newline e^{2x}-4e^x+1=0; \newline$

Lets $y=e^x; \newline$

$y^2-4y+1=0; \newline D=b^2-4ac; \newline D=16-4*1*1=12; \newline \sqrt{D}=\sqrt{12}=2\sqrt{3}; \newline y_1=(-b+\sqrt{D})/2a; \newline y_1=(4+2\sqrt{3})/2=2+\sqrt{3}=2+1.73=3.73; \newline y_2=(-b-\sqrt{D})/2a; \newline y_2=(4-2\sqrt{3})/2=2-\sqrt{3}=2-1.73=0.27; \newline$

The main natural logarithm formula is:

$ln(e^x)=x. \newline y_1=e^{x_1};\newline e^{x_1}=3.73;\newline ln(e^{x_1})=ln(3.73);\newline x_1*ln(e)=ln(3.73);\newline x_1*1=ln(3.73);\newline x_1=1.32; \newline y_2=e^{x_2}; \newline e^{x_2}=0.27; \newline ln(e^{x_2})=ln(0.27); \newline x_2=ln(0.27); \newline x_2=-1.3.$

Answer: $x_1=1.32; x_2=-1.3.$