
If \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] are in A.P then the values of \[{\rm{x}}\] are
A. \[{\log _5}4{\rm{ and }}{\log _5}3\]
B. \[{\log _3}4{\rm{ and }}{\log _4}3\]
C. \[{\log _3}4{\rm{ and }}{\log _3}5\]
D. \[{\log _5}6{\rm{ and }}{\log _5}7\]
Answer
161.1k+ views
Hint: We are provided the statement that, the terms of Arithmetic progression are \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] are in A.P. For that we have to write the given terms with respect to \[2b = a + c\] and then substituting the given terms in \[2b = a + c\] and solve accordingly using power rule of log to get the desired solution.
Formula Used: Product rule:
\[{\log _{\rm{a}}}({\rm{mn}}) = {\log _{\rm{a}}}{\rm{m}} + {\log _{\rm{a}}}{\rm{n}}\]
Power rule:
\[{\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{n}}{\log _{\rm{a}}}{\rm{m}}\]
Complete step by step solution: We have been provided in the question that,
The terms \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] are in A.P
And we are to determine the values of \[{\rm{x}}\]
Since, it is given that \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] is in Arithmetic progression
Now, we have to use formula of A.P to rewrite the given terms, we have the formula as
\[2a = b + c\]
Now, writing the terms given with respect to \[2a = b + c\] is
\[2{\log _e}\left( {{5^x} - 1} \right) = {\log _e}5 + {\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\]
By using log property, write the above expression as,
\[ \Rightarrow {\left( {{5^x} - 1} \right)^2} = 5\left( {{5^x} - \dfrac{{11}}{5}} \right)\]
Now, we have to simplify the left side of the equation using the formula \[{(a - b)^2}\] and right side by taking LCM, we get
\[ \Rightarrow {5^{2x}} + 1 - 2 \times {5^x} = 5 \times {5^x} - 11\]
Now, on simplifying the above expression by grouping the like terms, we get
\[ \Rightarrow {5^{2x}} - 7 \times {5^x} + 12 = 0\]
Now, on expanding the above equation, we get
\[ \Rightarrow {5^{2x}} - 4 \times {5^x} - 3 \times {5^x} + 12 = 0\]
Now, we have to write the above expression in terms of factors, we have
\[ \Rightarrow \left( {{5^x} - 4} \right)\left( {{5^x} - 3} \right) = 0\]
Now, we have to equate each term to zero, we get
\[ \Rightarrow {5^x} = 4,{5^x} = 3\]
Now, on taking log on both sides of the above equation we get
\[ \Rightarrow \log {5^x} = \log 4,\log {5^x} = \log 3\]
Now, rewrite the above expression using the property\[\log {a^b} = b\log a\] we have
\[x = {\log _5}4,x = {\log _5}3\]
Therefore, if \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] are in A.P then the values of \[{\rm{x}}\] are \[{\log _5}4{\rm{ and }}{\log _5}3\]
Option ‘A’ is correct
Note: We must approach problems of this nature cautiously because there are more formulas to remember in this one. To get the right answer, students should thoroughly understand logarithm properties and log rules including the quotient rule, product rule, and power rule. Using the incorrect formula in the incorrect step will give the incorrect solution.
Formula Used: Product rule:
\[{\log _{\rm{a}}}({\rm{mn}}) = {\log _{\rm{a}}}{\rm{m}} + {\log _{\rm{a}}}{\rm{n}}\]
Power rule:
\[{\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{n}}{\log _{\rm{a}}}{\rm{m}}\]
Complete step by step solution: We have been provided in the question that,
The terms \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] are in A.P
And we are to determine the values of \[{\rm{x}}\]
Since, it is given that \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] is in Arithmetic progression
Now, we have to use formula of A.P to rewrite the given terms, we have the formula as
\[2a = b + c\]
Now, writing the terms given with respect to \[2a = b + c\] is
\[2{\log _e}\left( {{5^x} - 1} \right) = {\log _e}5 + {\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\]
By using log property, write the above expression as,
\[ \Rightarrow {\left( {{5^x} - 1} \right)^2} = 5\left( {{5^x} - \dfrac{{11}}{5}} \right)\]
Now, we have to simplify the left side of the equation using the formula \[{(a - b)^2}\] and right side by taking LCM, we get
\[ \Rightarrow {5^{2x}} + 1 - 2 \times {5^x} = 5 \times {5^x} - 11\]
Now, on simplifying the above expression by grouping the like terms, we get
\[ \Rightarrow {5^{2x}} - 7 \times {5^x} + 12 = 0\]
Now, on expanding the above equation, we get
\[ \Rightarrow {5^{2x}} - 4 \times {5^x} - 3 \times {5^x} + 12 = 0\]
Now, we have to write the above expression in terms of factors, we have
\[ \Rightarrow \left( {{5^x} - 4} \right)\left( {{5^x} - 3} \right) = 0\]
Now, we have to equate each term to zero, we get
\[ \Rightarrow {5^x} = 4,{5^x} = 3\]
Now, on taking log on both sides of the above equation we get
\[ \Rightarrow \log {5^x} = \log 4,\log {5^x} = \log 3\]
Now, rewrite the above expression using the property\[\log {a^b} = b\log a\] we have
\[x = {\log _5}4,x = {\log _5}3\]
Therefore, if \[{\log _e}5,{\log _e}\left( {{5^x} - 1} \right)\] and \[{\log _e}\left( {{5^x} - \dfrac{{11}}{5}} \right)\] are in A.P then the values of \[{\rm{x}}\] are \[{\log _5}4{\rm{ and }}{\log _5}3\]
Option ‘A’ is correct
Note: We must approach problems of this nature cautiously because there are more formulas to remember in this one. To get the right answer, students should thoroughly understand logarithm properties and log rules including the quotient rule, product rule, and power rule. Using the incorrect formula in the incorrect step will give the incorrect solution.
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