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Sagot :
PART A
We are given the expression as
[tex]\ln (e^2\ln e^3)[/tex]We will use the laws of the logarithm to find the solution
We can open up the expression to give other expressions using the product law of logarithm, and log of power rule.
The product law is given as
[tex]\ln (M\times N)=lnM+lnN[/tex]The log of power rule is given as
[tex]\ln x^b=b\ln x[/tex]All these rules will be applied in the solution below. Also note the following
[tex]\begin{gathered} \ln e=1 \\ \ln 1=0 \end{gathered}[/tex]we can then solve the expression to give.
[tex]\begin{gathered} \ln (e^2\ln e^3) \\ =\ln (e\times e\times\ln e^3) \\ =\ln e+\ln e+\ln (\ln e^3) \\ \sin ce\text{ lne=1} \\ =1+1+\ln (\ln e^3) \\ =2+\ln (3\ln e) \\ =2+\ln (3\times\ln e) \\ =2+\ln 3+ln(\ln e) \\ =2+\ln 3+\ln 1 \\ \sin ce\text{ ln1=0} \\ =2+\ln 3 \end{gathered}[/tex]Therefore the answer to the first part is:
[tex]2+\ln 3[/tex]PART B
We are given the expression as
[tex]\begin{gathered} \ln (8x)^{\frac{1}{2}}+\ln 4x^2-\ln (16x)^{\frac{1}{2}} \\ \end{gathered}[/tex]In addition to the rules stated before, we will also make use of an additional rule known as the quotient rule of logarithm.
[tex]\ln \mleft(\frac{x}{y}\mright)=\ln \mleft(x\mright)-\ln \mleft(y\mright)[/tex]We can then find the solution of the given expression use all the rules.
[tex]\begin{gathered} \ln (8x)^{\frac{1}{2}}+\ln 4x^2-\ln (16x)^{\frac{1}{2}} \\ =\ln (\frac{(8x)^{\frac{1}{2}}\times4x^2}{(16x)^{\frac{1}{2}}}) \\ =\ln (\frac{8^{\frac{1}{2}}x^{\frac{1}{2}}\times4x^2}{16^{\frac{1}{2}}x^{\frac{1}{2}}}) \\ =\ln (\frac{8^{\frac{1}{2}}\times4x^2}{16^{\frac{1}{2}}}) \\ =\ln \frac{(2^{3\times\frac{1}{2}}\times4x^2)}{2^{4\times\frac{1}{2}}} \\ =\ln \frac{(2^{\frac{3}{2}}\times4x^2)}{2^2} \\ =\ln (\frac{(2^{\frac{3}{2}}\times4x^2)}{4}) \\ =\ln (2^{\frac{3}{2}}\times x^2) \\ =\ln (2\sqrt[]{2}\times x^2) \\ =\ln (2\sqrt[]{2}x^2) \end{gathered}[/tex]Therefore the answer for part B is
[tex]\ln (2\sqrt[]{2}x^2)[/tex]Note: You can test for the accuracy of the answer, by substituting any arbitrary value for x in both the question and answer, you would see that it gives the same value when using your calculator.
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