Question

How do you simplify \[\ln {{\left( 8x \right)}^{\dfrac{1}{2}}}+\ln 4{{x}^{2}}-\ln {{\left( 16x \right)}^{\dfrac{1}{2}}}\]?

Answer 1

Hint: In this problem, we have to simplify the given logarithmic expression. Here we are going to use some of the logarithmic identities to simplify the given expression. We can first use the logarithmic identity \[\ln a-\ln b=\ln \dfrac{a}{b}\] and simplify the expression, we can then further simplify the resulted expression with the identity \[\ln {{a}^{n}}=n\ln a\], we will then simplify the resulted expression by cross multiplying the terms with similar variable to get the final answer.

Complete step by step solution:
We know that the given expression to be simplified is,
\[\ln {{\left( 8x \right)}^{\dfrac{1}{2}}}+\ln 4{{x}^{2}}-\ln {{\left( 16x \right)}^{\dfrac{1}{2}}}\]
We can now simplify this expression using some of the logarithmic identities.
We can now apply one of the logarithmic identities \[\ln a-\ln b=\ln \dfrac{a}{b}\] to the given expression, we get
\[= \ln \left[ \dfrac{{{\left( 8x \right)}^{\dfrac{1}{2}}}+4{{x}^{2}}}{{{\left( 16x \right)}^{\dfrac{1}{2}}}} \right]\]
We can now simplify the above step, where we have half in the power, we can separate them into terms to be cancelled.
\[= \ln \left[ \dfrac{{{\left( 8 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}+4{{x}^{2}}}{{{\left( 16 \right)}^{\dfrac{1}{2}}}{{\left( x \right)}^{\dfrac{1}{2}}}} \right]\]
We can now cancel the similar terms and simplify the step further, we get
\[= \ln \left[ \dfrac{{{\left( 8 \right)}^{\dfrac{1}{2}}}+4{{x}^{2}}}{{{\left( 16 \right)}^{\dfrac{1}{2}}}} \right]=\ln \left[ \dfrac{{{2}^{2}}{{x}^{2}}}{{{2}^{\dfrac{1}{2}}}} \right]\]
We can now write the above step as,
\[= \ln \left[ {{2}^{2}}{{x}^{2}} \right]-\ln {{2}^{\dfrac{1}{2}}}\]
We can now use the identity \[\ln \left( ab \right)=\ln a+\ln b\] in the above step, we get
\[= \ln {{2}^{2}}+\ln {{x}^{2}}-\ln {{2}^{\dfrac{1}{2}}}\]
We can now apply the identity \[\ln {{a}^{n}}=n\ln a\] in the above step, we get
\[= 2\ln 2+2\ln x-\dfrac{1}{2}\ln 2\]
We can now simplify the above step, we get
\[= 2\ln x+\dfrac{3}{2}\ln 2\]
Therefore, the simplified form of the given expression \[\ln {{\left( 8x \right)}^{\dfrac{1}{2}}}+\ln 4{{x}^{2}}-\ln {{\left( 16x \right)}^{\dfrac{1}{2}}}\] is\[2\ln x+\dfrac{3}{2}\ln 2\].

Note: We should always remember the basic logarithmic formulas, rules and identities to be used in these types of problems to be simplified. We should remember the identities like \[\ln {{a}^{n}}=n\ln a\], \[\ln a-\ln b=\ln \dfrac{a}{b}\] and \[\ln \left( ab \right)=\ln a+\ln b\], which are the basic formulas.