Solution.
(e2x+1)+3−3(ex+1)=ex;(e2x+1)+3−3ex−3=ex;e2x−4ex+1=0;
Lets y=ex;
y2−4y+1=0;D=b2−4ac;D=16−4∗1∗1=12;D=12=23;y1=(−b+D)/2a;y1=(4+23)/2=2+3=2+1.73=3.73;y2=(−b−D)/2a;y2=(4−23)/2=2−3=2−1.73=0.27;
The main natural logarithm formula is:
ln(ex)=x.y1=ex1;ex1=3.73;ln(ex1)=ln(3.73);x1∗ln(e)=ln(3.73);x1∗1=ln(3.73);x1=1.32;y2=ex2;ex2=0.27;ln(ex2)=ln(0.27);x2=ln(0.27);x2=−1.3.
Answer: x1=1.32;x2=−1.3.
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