Question

Find \[{\log _{24}}48\] in terms of \[\alpha \] if \[{\log _{12}}36 = \alpha \].

Answer 1

Hint:
In this question, we are going to have to inverse the base and exponent in the second logarithm. Then we will have to equate the base and exponent of both the logarithms so as to solve for the given value and for expressing the first one in terms of \[\alpha \].

Formula Used:
In this question, there is going to be use of many formulae of basics of logarithms like splitting of an exponent of logarithms as two more, then we are also going to use inversion of logarithms:
\[{\log _a}b \times c\]= \[{\log _a}b\]+ \[{\log _a}c\], \[{\log _a}b\]=\[\dfrac{1}{{{{\log }_b}a}}\] and \[{\log _a}{b^n}\]=\[n{\log _a}b\]

Complete step by step solution:
We know that the basic formula of logarithm says that:
If \[{\log _a}b\]=\[c\], then \[{a^c}\]= b
So, we can say that logarithm is the inverse of exponentiation function.

Now, we are going to need to use some basic formulae of logarithm:
\[{\log _a}b \times c\]= \[{\log _a}b\]+ \[{\log _a}c\]
This formula defines the rule if you have to split a composite exponent inside a logarithm as the sum of its constituent factors.
\[{\log _a}b\]=\[\dfrac{1}{{{{\log }_b}a}}\]
This formula defines the rule if I have to inverse the base and exponent of a logarithm function.
\[{\log _a}{b^n}\]=\[n{\log _a}b\]
This formula defines the rule when the exponent is raised to a power.
And, \[{\log _a}a\]=\[1\], because \[{a^1}\]=\[a\]

Now, in this question, we are given \[{\log _{12}}36\]=\[\alpha \] .
Since \[{6^2}\]= \[36\], so
\[{\log _{12}}{6^2}\]= \[2{\log _{12}}6\] (from using the above formula)
So, \[2{\log _{12}}6\]=\[\alpha \]

Also, using the above formula, we can say that:
\[{\log _{12}}6\]=\[\dfrac{1}{{{{\log }_6}12}}\]
Hence, \[\alpha \]= \[2 \times \dfrac{1}{{{{\log }_6}12}}\]
or, \[\dfrac{\alpha }{2}\]=\[\dfrac{1}{{{{\log }_6}12}}\]
Now, \[{\log _6}12\]=\[{\log _6}6\]+\[{\log _6}2\]=\[1\]+\[\dfrac{1}{{{{\log }_2}6}}\]
Now, \[{\log _2}6\]=\[{\log _2}2\]+\[{\log _2}3\]=\[1\]+\[{\log _2}3\]
Hence, \[{\log _6}12\]=\[1\]+\[\dfrac{1}{{1 + {{\log }_2}3}}\]=\[\dfrac{{1 + {{\log }_2}3}}{{1 + {{\log }_2}3}}\]+\[\dfrac{1}{{1 + {{\log }_2}3}}\]=\[\dfrac{{1 + 1 + {{\log }_2}3}}{{1 + {{\log }_2}3}}\]=\[\dfrac{{2 + {{\log }_2}3}}{{1 + {{\log }_2}3}}\] (taking the LCM)
So, \[\dfrac{\alpha }{2}\]=\[\dfrac{{1 + {{\log }_2}3}}{{2 + {{\log }_2}3}}\]
And if we take the reciprocal on both sides, we get,
\[\dfrac{2}{\alpha }\]=\[\dfrac{{2 + {{\log }_2}3}}{{1 + {{\log }_2}3}}\]
Now, cross-multiplying the expressions on the two sides of the equality:
\[2 + 2{\log _2}3\]=\[2\alpha + \alpha {\log _2}3\]
Rearranging the terms, we get:
\[2{\log _2}3 - \alpha {\log _2}3 = 2\alpha - 2\]
Taking commons from the two sides:
\[{\log _2}3\left( {2 - \alpha } \right) = 2\left( {\alpha - 1} \right)\]
So, \[{\log _2}3 = \dfrac{{2\left( {\alpha - 1} \right)}}{{2 - \alpha }}\]

Now, \[{\log _{24}}48 = {\log _{24}}24 + {\log _{24}}2 = 1 + \dfrac{1}{{{{\log }_2}24}}\]
Now, \[{\log _2}24 = {\log _2}8 + {\log _2}3 = {\log _2}{2^3} + {\log _2}3 = 3 + {\log _2}3\]
Putting the value of \[{\log _2}3\]from above we have:
\[{\log _2}24 = 3 + \dfrac{{2\left( {\alpha - 1} \right)}}{{2 - \alpha }}\]
Taking the LCM and solving, we have:
\[{\log _2}24 = \dfrac{{6 - 3\alpha + 2\alpha - 2}}{{2 - \alpha }} = \dfrac{{4 - \alpha }}{{2 - \alpha }}\]
Substituting the value, we get:
\[{\log _{24}}48 = 1 + \dfrac{1}{{{{\log }_2}24}}\]
Now, putting the value of \[{\log _2}24 = \dfrac{{4 - \alpha }}{{2 - \alpha }}\], we get:
\[{\log _{24}}48 = 1 + \dfrac{1}{{\dfrac{{4 - \alpha }}{{2 - \alpha }}}}\]
\[{\log _{24}}48 = 1 + \dfrac{{2 - \alpha }}{{4 - \alpha }}\]
\[{\log _{24}}48 = \dfrac{{4 - \alpha + 2 - \alpha }}{{4 - \alpha }} = \dfrac{{6 - 2\alpha }}{{4 - \alpha }} = \dfrac{{2\left( {3 - \alpha } \right)}}{{4 - \alpha }}\]

Hence, the value of \[{\log _{24}}48 = \dfrac{{2\left( {3 - \alpha } \right)}}{{4 - \alpha }}\].

Note:
We saw that these questions totally rely on the student’s knowledge of application of logarithms and its ability to mold the given information in favor of the format given in the question. It is very important to write down all the steps in order as a mismatch leads to error, ruining the complete answer.