
If \[{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right),{\log _{10}}\left( {{2^x} + 3} \right)\] are three consecutive terms of an AP, then which one of the following is correct
A. \[x = 0\]
B. \[x = 1\]
C. \[x = {\log _2}5\]
D. \[x = {\log _5}2\]
Answer
160.8k+ views
Hint: In our case, we have been given the three consecutive terms of Arithmetic progression such as \[{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right),{\log _{10}}\left( {{2^x} + 3} \right)\] and now we have to determine the correct condition as per the given statement. 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:
\[{\log _{\rm{a}}}({\rm{mn}}) = {\log _{\rm{a}}}{\rm{m}} + {\log _{\rm{a}}}{\rm{n}}\]
Quotient rule:
\[{\log _{\rm{a}}}\dfrac{{\rm{m}}}{{\rm{n}}} = {\log _{\rm{a}}}{\rm{m}} - {\log _{\rm{a}}}{\rm{n}}\]
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 A.P's three consecutive terms are
\[{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right),{\log _{10}}\left( {{2^x} + 3} \right)\]
And we are asked to determine the condition that is correct according to the given statement.
We have been already known that if \[a,b,c\] are in A.P then
\[2b = a + c\]
Now, we have to wrote the given terms in terms of the above formula, we obtain
\[ \Rightarrow 2 \times {\log _{10}}\left( {{2^x} - 1} \right) = {\log _{10}}2 + {\log _{10}}\left( {{2^x} + 3} \right)\]
Now, using power rule (\[{\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{n}}{\log _{\rm{a}}}{\rm{m}}\]) we can write the above equation as
\[ \Rightarrow {\log _{10}}{\left( {{2^x} - 1} \right)^2} = {\log _{10}}\left( {2 \times \left( {{2^x} + 3} \right)} \right)\]
Now, after neglecting log, we have
\[ \Rightarrow {\left( {{2^x} - 1} \right)^2} = 2 \times \left( {{2^x} + 3} \right)\]---- (1)
Now, for instance let us assume
\[{2^x} = t\]
Now, on rewriting the equation (1) in terms of above assumption \[{2^x} = t\] we get
\[ \Rightarrow {\left( {t - 1} \right)^2} = 2 \times \left( {t + 3} \right)\]
Now using \[{\left( {a - b} \right)^2}\] formula, expand the left side of the above equation we get
\[ \Rightarrow {t^2} - 2t + 1 = 2(t + 3)\]
Now, we have to solve the right side of the equation by multiplying 2 with the terms inside the parentheses we have
\[ \Rightarrow {t^2} - 2t + 1 = 2t + 6\]
Now, we have to group the like terms to simplify, we get
\[ \Rightarrow {t^2} - 4t - 5 = 0\]
Now, we have to write the above expression in terms of factors, we obtain
\[ \Rightarrow (t - 5)(t + 1) = 0\]
Now, we have to set factors equal to zero, we get
\[t + 1 = 0\;\]OR \[t - 5 = 0\]
On soling for t, we get
\[t = - 1\] OR \[t = 5\]
Since, \[{2^x} > 0\] now it implies,
\[{2^x} \ne - 1\]
Now, we have that
\[{2^x} = 5\]
Now, we have to write in terms of log for that take log on both sides, we get
\[{\log _2}x = {\log _2}5\]
Now, rewrite the above expression using the property\[\log {a^b} = b\log a\] we have
\[x{\log _2}2 = {\log _2}5\]
We know that \[{\log _2}2 = 1\] so, we have
\[x = {\log _2}5\]
Therefore, if \[{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right),{\log _{10}}\left( {{2^x} + 3} \right)\] are three consecutive terms of an AP, then \[x = {\log _2}5\] is correct
Option ‘C’ is correct
Note: As this problem has more formulas to be remembered, we have to solve these types of problems carefully. Students should be thorough with properties of logarithm and log rules like quotient rule, product rule and power rule in order to get the correct solution. Because applying wrong formula in wrong step leads to wrong solution.
Formula Used: Product rule:
\[{\log _{\rm{a}}}({\rm{mn}}) = {\log _{\rm{a}}}{\rm{m}} + {\log _{\rm{a}}}{\rm{n}}\]
Quotient rule:
\[{\log _{\rm{a}}}\dfrac{{\rm{m}}}{{\rm{n}}} = {\log _{\rm{a}}}{\rm{m}} - {\log _{\rm{a}}}{\rm{n}}\]
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 A.P's three consecutive terms are
\[{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right),{\log _{10}}\left( {{2^x} + 3} \right)\]
And we are asked to determine the condition that is correct according to the given statement.
We have been already known that if \[a,b,c\] are in A.P then
\[2b = a + c\]
Now, we have to wrote the given terms in terms of the above formula, we obtain
\[ \Rightarrow 2 \times {\log _{10}}\left( {{2^x} - 1} \right) = {\log _{10}}2 + {\log _{10}}\left( {{2^x} + 3} \right)\]
Now, using power rule (\[{\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{n}}{\log _{\rm{a}}}{\rm{m}}\]) we can write the above equation as
\[ \Rightarrow {\log _{10}}{\left( {{2^x} - 1} \right)^2} = {\log _{10}}\left( {2 \times \left( {{2^x} + 3} \right)} \right)\]
Now, after neglecting log, we have
\[ \Rightarrow {\left( {{2^x} - 1} \right)^2} = 2 \times \left( {{2^x} + 3} \right)\]---- (1)
Now, for instance let us assume
\[{2^x} = t\]
Now, on rewriting the equation (1) in terms of above assumption \[{2^x} = t\] we get
\[ \Rightarrow {\left( {t - 1} \right)^2} = 2 \times \left( {t + 3} \right)\]
Now using \[{\left( {a - b} \right)^2}\] formula, expand the left side of the above equation we get
\[ \Rightarrow {t^2} - 2t + 1 = 2(t + 3)\]
Now, we have to solve the right side of the equation by multiplying 2 with the terms inside the parentheses we have
\[ \Rightarrow {t^2} - 2t + 1 = 2t + 6\]
Now, we have to group the like terms to simplify, we get
\[ \Rightarrow {t^2} - 4t - 5 = 0\]
Now, we have to write the above expression in terms of factors, we obtain
\[ \Rightarrow (t - 5)(t + 1) = 0\]
Now, we have to set factors equal to zero, we get
\[t + 1 = 0\;\]OR \[t - 5 = 0\]
On soling for t, we get
\[t = - 1\] OR \[t = 5\]
Since, \[{2^x} > 0\] now it implies,
\[{2^x} \ne - 1\]
Now, we have that
\[{2^x} = 5\]
Now, we have to write in terms of log for that take log on both sides, we get
\[{\log _2}x = {\log _2}5\]
Now, rewrite the above expression using the property\[\log {a^b} = b\log a\] we have
\[x{\log _2}2 = {\log _2}5\]
We know that \[{\log _2}2 = 1\] so, we have
\[x = {\log _2}5\]
Therefore, if \[{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right),{\log _{10}}\left( {{2^x} + 3} \right)\] are three consecutive terms of an AP, then \[x = {\log _2}5\] is correct
Option ‘C’ is correct
Note: As this problem has more formulas to be remembered, we have to solve these types of problems carefully. Students should be thorough with properties of logarithm and log rules like quotient rule, product rule and power rule in order to get the correct solution. Because applying wrong formula in wrong step leads to wrong solution.
Recently Updated Pages
Crack JEE Advanced 2025 with Vedantu's Live Classes

JEE Advanced Maths Revision Notes

JEE Advanced Chemistry Revision Notes

Download Free JEE Advanced Revision Notes PDF Online for 2025

The students S1 S2 S10 are to be divided into 3 groups class 11 maths JEE_Advanced

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

JEE Advanced Cut Off 2024

JEE Advanced Exam Pattern 2025

Other Pages
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
