(i)
e2x⋅(e2x−1)3=9 Use substitution e2x=t,t>0. Then
t(t−1)3=9
t4−3t3+3t2−t−9=0
t>0=>t≈2.527
e2x≈2.527
x≈0.5ln2.527
x≈0.464
(ii)
2lnx−ln(3−5x)=ln2 x>0,3−5x>0
Then 0<x<53
lnx2=ln(2(3−5x))
x2=6−10x
x2+10x−6=0
D=(10)2−4(1)(−6)=124
x=2(1)−10±124=−5±31
−5+31<53
31<528
31<25784
31<31.36,True
Answer: x=31−5.
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