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Question

Solve the following equations.
(6x5)|ln(2x2.3)|=8ln(2x+2.3)

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Solution

(6x5)|ln(2x2.3)|=8ln(2x+2.3)
We know that ln x>0x>1
and lnx<00<x<1
(6x5)|ln(2x+2.3)|=8ln(2x+2.3)
case I :
If 2x+2.3>1
x1.3/2
x13/20, then
(6x5)ln(2x+2.3)8ln(2x+2.3)=0
ln(2x+2.3)[6x13]=0
ln(2x+2.3)=0 or 6x13=0
2x+2.3=1 or x=13/6
x=13/20 or x=13/6
case 2 :
if 2x+2.3<1x<13/20 then
(6x5)ln(2x+2.3)8ln(2x+2.3)=0
ln(2x+2.3)(6x+3)=0
ln(2x+2.3)=0 = or 6x+3=0
2x+2.3=1 or x=1/2
x=13/20
However in the case 2 the allowed region is a<13/20
Hence x=13/20 & 1/2 are not permissible
x=13/20 & x=13/6 are the only solution possible


1137955_887982_ans_9be58a8f9b2c4fd18cd347afd556b5b9.jpg

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