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91Ó°ÊÓ

Write each expression as the sum and/or difference of logarithms. Express powers as factors.

ln2x+3x2-3x+22,x>2

Short Answer

Expert verified

ln2x+3x2-3x+22=2ln2x+3-2lnx-2-2lnx-1

Step by step solution

01

Given data

The expression is :ln2x+3x2-3x+22,x>2

02

Concept used

Properties of logarithms:

  • logaMN=logaM+logaN
  • logaMN=logaM-logaN
  • logaMr=rlogaM
  • ax=exlna
03

Evaluate 

ln2x+3x2-3x+22=2ln2x+3x2-3x+2; Using property of logarithms

ln2x+3x2-3x+22=2ln2x+3-lnx2-3x+2; Using property of logarithms

role="math" localid="1646483367708" ln2x+3x2-3x+22=2ln2x+3-2lnx2-3x+2; Using distributive law

ln2x+3x2-3x+22=2ln2x+3-2lnx-2x-1; By factorization

ln2x+3x2-3x+22=2ln2x+3-2lnx-2+lnx-1; Using property of logarithms

role="math" localid="1646483563558" ln2x+3x2-3x+22=2ln2x+3-2lnx-2-2lnx-1 ; By distributive law

04

Conclusion

Thus,ln2x+3x2-3x+22=2ln2x+3-2lnx-2-2lnx-1

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