Answer
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Hint: Use logarithmic rules to solve this problem.
1.\[\log {}_ba = \dfrac{{\log a}}{{\log b}}\]
2.\[\log (a \times b) = \log a + \log b\]
3.\[\log {a^x} = x\log a\]
Complete step-by-step answer:
Given that, \[a = \log {}_23,\] \[b = \log {}_25\] and \[c = \log {}_72\]
We will use the first rule from the hint. Then
\[a = \dfrac{{\log 3}}{{\log 2}}\], \[b = \dfrac{{\log 5}}{{\log 2}}\] and \[c = \dfrac{{\log 2}}{{\log 7}}\] .
Now, moving towards the value we have to find
\[\log {}_{140}63\]
=\[\dfrac{{\log 63}}{{\log 140}}\] Using \[\log {}_ba = \dfrac{{\log a}}{{\log b}}\]
=\[\dfrac{{\log (9 \times 7)}}{{\log (2 \times 70)}}\] here factors should be used according to data given.
= \[\dfrac{{\log 9 + \log 7}}{{\log 2 + \log 70}}\] using \[\log (a \times b) = \log a + \log b\]
=\[\dfrac{{\log 9 + \log 7}}{{\log 2 + \log (2 \times 5 \times 7)}}\] Factorize number 70.
=\[\dfrac{{\log {3^2} + \log 7}}{{\log 2 + \log 2 + \log 5 + \log 7}}\] 9 can be written as square of 3 .
=\[\dfrac{{2\log 3 + \log 7}}{{2\log 2 + \log 5 + \log 7}}\] using \[\log {a^x} = x\log a\]
=\[\dfrac{{2a\log 2 + \dfrac{{\log 2}}{c}}}{{2\log 2 + b\log 2 + \dfrac{{\log 2}}{c}}}\] \[\log 3 = a\log 2,\log 5 = b\log 2,\log 7 = \dfrac{{\log 2}}{c}\]
rearranging the log terms so that all come in log2 form
=\[\dfrac{{\dfrac{{c \times 2a\log 2 + \log 2}}{c}}}{{\dfrac{{c \times 2\log 2 + bc \times \log 2 + \log 2}}{c}}}\] taking LCM separately for numerator and denominator.
=\[\dfrac{{\log 2(2ac + 1)}}{{\log 2(2c + bc + 1)}}\] cancelling C and taking log2 common.
=\[\dfrac{{2ac + 1}}{{2c + bc + 1}}\]
Thus,
\[\log {}_{140}63\]=\[\dfrac{{2ac + 1}}{{2c + bc + 1}}\]
Option d is the correct answer.
Note: When we solve problems related to logarithm we should try to simplify the ratios as much as we can.
If there is a need to split a number ,we should factorize it using the numbers in the given question.[like 63 and 140]. It is to make your calculations easy.
Don’t miss even a single step like finding L.C.M. or taking common terms.
1.\[\log {}_ba = \dfrac{{\log a}}{{\log b}}\]
2.\[\log (a \times b) = \log a + \log b\]
3.\[\log {a^x} = x\log a\]
Complete step-by-step answer:
Given that, \[a = \log {}_23,\] \[b = \log {}_25\] and \[c = \log {}_72\]
We will use the first rule from the hint. Then
\[a = \dfrac{{\log 3}}{{\log 2}}\], \[b = \dfrac{{\log 5}}{{\log 2}}\] and \[c = \dfrac{{\log 2}}{{\log 7}}\] .
Now, moving towards the value we have to find
\[\log {}_{140}63\]
=\[\dfrac{{\log 63}}{{\log 140}}\] Using \[\log {}_ba = \dfrac{{\log a}}{{\log b}}\]
=\[\dfrac{{\log (9 \times 7)}}{{\log (2 \times 70)}}\] here factors should be used according to data given.
= \[\dfrac{{\log 9 + \log 7}}{{\log 2 + \log 70}}\] using \[\log (a \times b) = \log a + \log b\]
=\[\dfrac{{\log 9 + \log 7}}{{\log 2 + \log (2 \times 5 \times 7)}}\] Factorize number 70.
=\[\dfrac{{\log {3^2} + \log 7}}{{\log 2 + \log 2 + \log 5 + \log 7}}\] 9 can be written as square of 3 .
=\[\dfrac{{2\log 3 + \log 7}}{{2\log 2 + \log 5 + \log 7}}\] using \[\log {a^x} = x\log a\]
=\[\dfrac{{2a\log 2 + \dfrac{{\log 2}}{c}}}{{2\log 2 + b\log 2 + \dfrac{{\log 2}}{c}}}\] \[\log 3 = a\log 2,\log 5 = b\log 2,\log 7 = \dfrac{{\log 2}}{c}\]
rearranging the log terms so that all come in log2 form
=\[\dfrac{{\dfrac{{c \times 2a\log 2 + \log 2}}{c}}}{{\dfrac{{c \times 2\log 2 + bc \times \log 2 + \log 2}}{c}}}\] taking LCM separately for numerator and denominator.
=\[\dfrac{{\log 2(2ac + 1)}}{{\log 2(2c + bc + 1)}}\] cancelling C and taking log2 common.
=\[\dfrac{{2ac + 1}}{{2c + bc + 1}}\]
Thus,
\[\log {}_{140}63\]=\[\dfrac{{2ac + 1}}{{2c + bc + 1}}\]
Option d is the correct answer.
Note: When we solve problems related to logarithm we should try to simplify the ratios as much as we can.
If there is a need to split a number ,we should factorize it using the numbers in the given question.[like 63 and 140]. It is to make your calculations easy.
Don’t miss even a single step like finding L.C.M. or taking common terms.
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