Answer
Verified
395.1k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE