Asked by Byron Sherwin on May 10, 2024
Verified
Use the properties of logarithms to expand ln2xx−6\ln \sqrt { \frac { 2 x } { x - 6 } }lnx−62x .
A) (ln2+lnx−ln(x−6) ) 1/2( \ln 2 + \ln x - \ln ( x - 6 ) ) ^ { 1 / 2 }(ln2+lnx−ln(x−6) ) 1/2
B) (ln2⋅lnxlnx−ln6) 1/2\left( \frac { \ln 2 \cdot \ln x } { \ln x - \ln 6 } \right) ^ { 1 / 2 }(lnx−ln6ln2⋅lnx) 1/2
C) 12(ln2+lnx−lnxln6) \frac { 1 } { 2 } \left( \ln 2 + \ln x - \frac { \ln x } { \ln 6 } \right) 21(ln2+lnx−ln6lnx)
D) 12(ln2+lnx−ln(x−6) ) \frac { 1 } { 2 } ( \ln 2 + \ln x - \ln ( x - 6 ) ) 21(ln2+lnx−ln(x−6) )
E) 12ln2+lnx−ln(x−6) \frac { 1 } { 2 } \ln 2 + \ln x - \ln ( x - 6 ) 21ln2+lnx−ln(x−6)
Properties
The characteristics, attributes or qualities that an object, substance, or mathematical entity possesses.
Logarithms
The exponents to which a fixed base must be raised to produce a given number.
Expand
In mathematics, expanding refers to the process of simplifying an expression or equation by multiplying out brackets and combining like terms.
- Rely on logarithmic properties for the operations of expansion and condensation.
Verified Answer
MN
Minh-Tu NguyenMay 14, 2024
Final Answer :
D
Explanation :
We can use the properties of logarithms to simplify the given expression:
ln2xx−6=12ln(2xx−6)=12(ln2x−ln(x−6))=12ln2+12lnx−12ln(x−6)=12(ln2+lnx−ln(x−6))\begin{align*}\ln \sqrt{\frac{2x}{x-6}} &= \frac{1}{2} \ln \left(\frac{2x}{x-6}\right) \\&= \frac{1}{2} \left(\ln 2x - \ln (x-6)\right) \\&= \frac{1}{2} \ln 2 + \frac{1}{2} \ln x - \frac{1}{2} \ln (x-6) \\&= \frac{1}{2} (\ln 2 + \ln x - \ln(x-6))\end{align*}lnx−62x=21ln(x−62x)=21(ln2x−ln(x−6))=21ln2+21lnx−21ln(x−6)=21(ln2+lnx−ln(x−6))
So, the correct answer is D.
ln2xx−6=12ln(2xx−6)=12(ln2x−ln(x−6))=12ln2+12lnx−12ln(x−6)=12(ln2+lnx−ln(x−6))\begin{align*}\ln \sqrt{\frac{2x}{x-6}} &= \frac{1}{2} \ln \left(\frac{2x}{x-6}\right) \\&= \frac{1}{2} \left(\ln 2x - \ln (x-6)\right) \\&= \frac{1}{2} \ln 2 + \frac{1}{2} \ln x - \frac{1}{2} \ln (x-6) \\&= \frac{1}{2} (\ln 2 + \ln x - \ln(x-6))\end{align*}lnx−62x=21ln(x−62x)=21(ln2x−ln(x−6))=21ln2+21lnx−21ln(x−6)=21(ln2+lnx−ln(x−6))
So, the correct answer is D.
Learning Objectives
- Rely on logarithmic properties for the operations of expansion and condensation.
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