Answer:
[tex]\ln(\frac{xx + yy}{zz})^{\frac{1}{2}}[/tex]
Step-by-step explanation:
Given:
1/2(ln(xx + yy) − ln(zz))
Now,
From the properties of log function,
1) n × ln(x) = ln(xⁿ)
and,
2) ln(A) - ln(B) = [tex]\ln\frac{A}{B}[/tex]
applying the properties in the given equation
we get the above equation as:
[tex]\frac{1}{2}(\ln\frac{xx + yy}{zz})[/tex]
( using the property 2 we get (ln(xx + yy) − ln(zz) = [tex]\ln\frac{xx + yy}{zz}[/tex]
or
⇒ [tex]\ln(\frac{xx + yy}{zz})^{\frac{1}{2}}[/tex] ( using the property 1 i.e n × ln(x) = ln(xⁿ) )
expression as an equivalent expression with a single logarithm is [tex]\ln(\frac{xx + yy}{zz})^{\frac{1}{2}}[/tex]
