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If ${\log _{175}}5{\text{x = lo}}{{\text{g}}_{343}}7x$, then find the value of ${\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right)$ is
A)$1$
B)$2$
C)$3$
D)$4$
Answer
![Verified]()
Verified
Verified
Hint: We know that ${\log _e} a = k$ can also be written as$a = {e^k}$ So use this formula to solve given function and take their ratios to find the value of x. Put the value of x in ${\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right)$ and simplify to get the answer.
Complete step by step answer:
Given that ${\log _{175}}5{\text{x = lo}}{{\text{g}}_{343}}7x$ --- (i)
Let eq. (i) be equal to a constant k, then
$ \Rightarrow {\log _{175}}5{\text{x = lo}}{{\text{g}}_{343}}7x = k$
We have to find the value of ${\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right)$.
Now we know that ${\log _e}a = k$ can also be written as $a = {e^k}$ so using this we get,
$ \Rightarrow {\log _{175}}5{\text{x = k}} \Rightarrow {\text{5x = 17}}{{\text{5}}^k}$ --- (ii)
And \[\; \Rightarrow {\text{lo}}{{\text{g}}_{343}}7x = k \Rightarrow 7x = {343^k}\] --- (iii)
On taking the ratio of eq. (ii) and (iii), we get
$ \Rightarrow \dfrac{{5x}}{{7x}} = \dfrac{{{{175}^k}}}{{{{343}^k}}}$
On canceling x we get,
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{175}}{{343}}} \right)^k}$
Now we can break $175$ and $343$ into factors. We can write $175 = 5 \times 5 \times 7$ and $343 = 7 \times 7 \times 7$.
On putting these factors in the equation we get,
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{5 \times 5 \times 7}}{{7 \times 7 \times 7}}} \right)^k}$
On cancelling $7$ from numerator and denominator, we get-
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{5 \times 5}}{{7 \times 7}}} \right)^k}$
We can also write it as-
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{{5^2}}}{{{7^2}}}} \right)^k}$
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{5}{7}} \right)^{2k}}$
Since the given ratios are equal then powers of the given ratios will also be equal.
$ \Rightarrow 1 = 2k$
On transferring $2$ from right to left we get,
$ \Rightarrow \dfrac{1}{2} = k$
Now we know the value of k. So on putting this value in eq. (ii) we get,
$ \Rightarrow 5x = {175^{\dfrac{1}{2}}}$
We can write$175 = 5 \times 5 \times 7$ then we get,
$ \Rightarrow 5x = {\left( {5 \times 5 \times 7} \right)^{\dfrac{1}{2}}}$
$ \Rightarrow 5x = {\left( {{5^2} \times 7} \right)^{\dfrac{1}{2}}}$
On simplifying we get,
$ \Rightarrow 5x = 5\sqrt 7 \Rightarrow x = \sqrt 7 $
Now we know the value of x. On putting the value of x in ${\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right)$, we get
$ \Rightarrow {\log _{42}}\left( {{{\left( {{7^{\dfrac{1}{2}}}} \right)}^4} - 2{{\left( {{7^{\dfrac{1}{2}}}} \right)}^2} + 7} \right)$
On simplifying we get,
$ \Rightarrow {\log _{42}}\left( {{7^2} - 2 \times 7 + 7} \right)$
$ \Rightarrow {\log _{42}}\left( {49 - 14 + 7} \right)$
$ \Rightarrow {\log _{42}}\left( {35 + 7} \right)$
$ \Rightarrow {\log _{42}}42$
We know that ${\log _e}e = 1$ then
$ \Rightarrow {\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right) = 1$
Hence, the correct option is ‘A’.
Note: John Napier introduced the concept of logarithms which was later used by scientists and engineers to make the calculations simple. The logarithm is inverse of the exponential. It is used in surveying and celestial navigation. It is used to measure the loudness of sound in decibels, the intensity of the earthquake, and in radioactive decay. Remember that if bases are equal exponents will also be equal.
Complete step by step answer:
Given that ${\log _{175}}5{\text{x = lo}}{{\text{g}}_{343}}7x$ --- (i)
Let eq. (i) be equal to a constant k, then
$ \Rightarrow {\log _{175}}5{\text{x = lo}}{{\text{g}}_{343}}7x = k$
We have to find the value of ${\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right)$.
Now we know that ${\log _e}a = k$ can also be written as $a = {e^k}$ so using this we get,
$ \Rightarrow {\log _{175}}5{\text{x = k}} \Rightarrow {\text{5x = 17}}{{\text{5}}^k}$ --- (ii)
And \[\; \Rightarrow {\text{lo}}{{\text{g}}_{343}}7x = k \Rightarrow 7x = {343^k}\] --- (iii)
On taking the ratio of eq. (ii) and (iii), we get
$ \Rightarrow \dfrac{{5x}}{{7x}} = \dfrac{{{{175}^k}}}{{{{343}^k}}}$
On canceling x we get,
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{175}}{{343}}} \right)^k}$
Now we can break $175$ and $343$ into factors. We can write $175 = 5 \times 5 \times 7$ and $343 = 7 \times 7 \times 7$.
On putting these factors in the equation we get,
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{5 \times 5 \times 7}}{{7 \times 7 \times 7}}} \right)^k}$
On cancelling $7$ from numerator and denominator, we get-
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{5 \times 5}}{{7 \times 7}}} \right)^k}$
We can also write it as-
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{{{5^2}}}{{{7^2}}}} \right)^k}$
$ \Rightarrow \dfrac{5}{7} = {\left( {\dfrac{5}{7}} \right)^{2k}}$
Since the given ratios are equal then powers of the given ratios will also be equal.
$ \Rightarrow 1 = 2k$
On transferring $2$ from right to left we get,
$ \Rightarrow \dfrac{1}{2} = k$
Now we know the value of k. So on putting this value in eq. (ii) we get,
$ \Rightarrow 5x = {175^{\dfrac{1}{2}}}$
We can write$175 = 5 \times 5 \times 7$ then we get,
$ \Rightarrow 5x = {\left( {5 \times 5 \times 7} \right)^{\dfrac{1}{2}}}$
$ \Rightarrow 5x = {\left( {{5^2} \times 7} \right)^{\dfrac{1}{2}}}$
On simplifying we get,
$ \Rightarrow 5x = 5\sqrt 7 \Rightarrow x = \sqrt 7 $
Now we know the value of x. On putting the value of x in ${\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right)$, we get
$ \Rightarrow {\log _{42}}\left( {{{\left( {{7^{\dfrac{1}{2}}}} \right)}^4} - 2{{\left( {{7^{\dfrac{1}{2}}}} \right)}^2} + 7} \right)$
On simplifying we get,
$ \Rightarrow {\log _{42}}\left( {{7^2} - 2 \times 7 + 7} \right)$
$ \Rightarrow {\log _{42}}\left( {49 - 14 + 7} \right)$
$ \Rightarrow {\log _{42}}\left( {35 + 7} \right)$
$ \Rightarrow {\log _{42}}42$
We know that ${\log _e}e = 1$ then
$ \Rightarrow {\log _{42}}\left( {{{\text{x}}^4} - 2{{\text{x}}^2} + 7} \right) = 1$
Hence, the correct option is ‘A’.
Note: John Napier introduced the concept of logarithms which was later used by scientists and engineers to make the calculations simple. The logarithm is inverse of the exponential. It is used in surveying and celestial navigation. It is used to measure the loudness of sound in decibels, the intensity of the earthquake, and in radioactive decay. Remember that if bases are equal exponents will also be equal.
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