Write the expression as a logarithm of a single quantity. 2[\ln x-\ln(x+

cleritere39

cleritere39

Answered question

2021-11-17

Write the expression as a logarithm of a single quantity.
2[lnxln(x+1)ln(x1)]

Answer & Explanation

Charles Randolph

Charles Randolph

Beginner2021-11-18Added 16 answers

Step 1
Given:
Let y=2[lnxln(x+1)ln(x1)]
To Find: To write the given expression as a logarithm of a single quantity.
Step 2
Solution:
y=2[lnxln(x+1)ln(x1)]
2[lnx[ln(x+1)ln(x1)]]
We know that  lna+lnb=ln(ab)    (Properties of logarithms)
y=2[lnxln[(x+1)(x1)]]
2[lnxln(x21)]      [Using  (a+b)(ab)=a2b2]
We know that  lnalnb=lnab   (Properties of logarithms)
y=2ln(xx21)=ln(xx21)2  (Since  mlna=lnam)
Hence  2[lnxln(x+1)ln(x1)]=ln(xx21)2
Harr1957

Harr1957

Beginner2021-11-19Added 18 answers

2[lnxln(x+1)ln(x1)]
2[lnxln(x+1)ln(x1)]=2[ln(xx+1)ln(x1)]
(the property  lnalnb=lnab  treat  x  as  a  and  (x+1)  as  b)
=2ln(x(x+1)x1)
(the property  lnalnb=lnab  treat  x(x+1)  as  a  and  (x1)  as  b)
=ln(x(x+1)x1)2
(the property  blna=ln(ab))
=ln(x(x1)(x+1))2(  simplify)
Result:ln(xx21)

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