If log_3 2, log_3(2^x - 5) and log_3(2^x - 7/ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $\log_3 2, \log_3(2^x - 5)$,and $\log_3(2^x - 7/2)$ are in $A.P.$Since they are in $A.P.$,the middle term is the arithmetic mean of the

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given that $\log_3 2, \log_3(2^x - 5)$,and $\log_3(2^x - 7/2)$ are in $A.P.$Since they are in $A.P.$,the middle term is the arithmetic mean of the other two: $2 \log_3(2^x - 5) = \log_3 2 + \log_3(2^x - 7/2)$.Using the property $\log_b a + \log_b c = \log_b(ac)$,we get: $\log_3(2^x - 5)^2 = \log_3[2(2^x - 7/2)]$.Equating the arguments: $(2^x - 5)^2 = 2(2^x) - 7$.Let $y = 2^x$. Then $(y - 5)^2 = 2y - 7$.$y^2 - 10y + 25 = 2y - 7$.$y^2 - 12y + 32 = 0$.Factoring the quadratic: $(y - 8)(y - 4) = 0$,so $y = 8$ or $y = 4$.If $y = 8$,then $2^x = 8 \Rightarrow x = 3$.If $y = 4$,then $2^x = 4 \Rightarrow x = 2$.Check the domain: For $\log_3(2^x - 5)$ to be defined,$2^x - 5 > 0$,so $2^x > 5$.If $x = 2$,$2^2 = 4$,which is not $> 5$. Thus,$x = 2$ is rejected.If $x = 3$,$2^3 = 8$,which is $> 5$. Thus,$x = 3$ is the only solution.Since $3$ is not listed in options $A, B,$ or $C$,the correct option is $D$.

If $\log_3 2, \log_3(2^x - 5)$ and $\log_3(2^x - 7/2)$ are in $A.P.$,then $x$ is equal to

Similar Questions

The sum of the $3^{rd}$ and the $4^{th}$ terms of a $G.P.$ is $60$ and the product of its first three terms is $1000$. If the first term of this $G.P.$ is positive,then its $7^{th}$ term is

If $\sum_{r=1}^{n}r^3 - \sum_{p=1}^{n} \sum_{m=1}^{p} \sum_{r=1}^{m} 1 = 80$,then the possible value of $n$ is:

$(1^{2}+2^{2}+3^{2}+\cdots+10^{2})$ is equal to:

If $A = 1 + r^z + r^{2z} + r^{3z} + .......\infty$,then the value of $r$ will be

The value of $\frac{C_1}{C_0} + 2 \cdot \frac{C_2}{C_1} + 3 \cdot \frac{C_3}{C_2} + \dots + n \cdot \frac{C_n}{C_{n-1}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module