
Find the value of ${\log _9}27 - {\log _{27}}9$
Answer
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Hint: Before attempting this question, one should have prior knowledge about the properties of logarithmic and make sure you try to make bases the same of all logarithmic functions in the given equation which will help you to simplify the equation.
Complete step-by-step answer:
According to the given information we have function ${\log _9}27 - {\log _{27}}9$
Let I = ${\log _9}27 - {\log _{27}}9$ (equation 1)
Now equation 1 can be written as ${\log _9}{\left( 3 \right)^3} - {\log _{27}}{\left( 3 \right)^2}$
Since we know that ${\log _a}{u^n} = n{\log _a}u$
Therefore, I = $3{\log _9}3 - 2{\log _{27}}3$
The above equation can be written as
I = \[3{\log _{{{\left( 3 \right)}^2}}}3 - 2{\log _{{{\left( 3 \right)}^3}}}3\]
Since we know that ${\log _a}u = \dfrac{1}{n}{\log _a}u$so after applying this property in the above equation we get
I = \[\dfrac{3}{2}{\log _3}3 - \dfrac{2}{3}{\log _3}3\]
Also, we know that ${\log _a}a = 1$ so applying in the above equation we get
I = $\dfrac{3}{2} - \dfrac{2}{3} = \dfrac{5}{6}$
Therefore, ${\log _9}27 - {\log _{27}}9 = \dfrac{5}{6}$
Note: In the above solution we came across the term logarithm which is an inverse function where inverse function can be defined as the functions which are reversible into another function. If f and g are the inverse functions then f(x) = y if and only if g(y) = x for example ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = {\cos ^{ - 1}}\left( {\cos {{60}^ \circ }} \right) = {60^ \circ }$ so $\cos {60^ \circ }$will be equal to 1.
Complete step-by-step answer:
According to the given information we have function ${\log _9}27 - {\log _{27}}9$
Let I = ${\log _9}27 - {\log _{27}}9$ (equation 1)
Now equation 1 can be written as ${\log _9}{\left( 3 \right)^3} - {\log _{27}}{\left( 3 \right)^2}$
Since we know that ${\log _a}{u^n} = n{\log _a}u$
Therefore, I = $3{\log _9}3 - 2{\log _{27}}3$
The above equation can be written as
I = \[3{\log _{{{\left( 3 \right)}^2}}}3 - 2{\log _{{{\left( 3 \right)}^3}}}3\]
Since we know that ${\log _a}u = \dfrac{1}{n}{\log _a}u$so after applying this property in the above equation we get
I = \[\dfrac{3}{2}{\log _3}3 - \dfrac{2}{3}{\log _3}3\]
Also, we know that ${\log _a}a = 1$ so applying in the above equation we get
I = $\dfrac{3}{2} - \dfrac{2}{3} = \dfrac{5}{6}$
Therefore, ${\log _9}27 - {\log _{27}}9 = \dfrac{5}{6}$
Note: In the above solution we came across the term logarithm which is an inverse function where inverse function can be defined as the functions which are reversible into another function. If f and g are the inverse functions then f(x) = y if and only if g(y) = x for example ${\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) = {\cos ^{ - 1}}\left( {\cos {{60}^ \circ }} \right) = {60^ \circ }$ so $\cos {60^ \circ }$will be equal to 1.
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