Here is a question from my book:
Given a GP and an AP with positive terms $a,a_1,a_2,a_3...a_n$ and $b,b_1,b_2,b_3,...b_n$ respectively. The common ratio of the GP is different from $1$.
Then there exists $x\in \Bbb R^+$, such that $\log_xa_n-\log_xa$ is equal to(A) $a-b$
(B) $a_n-b$
(C) $b_n-b$
(D) $a_n-b_n$
I asked for the solution from my teacher but he said that the question was wrong since there is no relation given between AP and GP. I told him that we can adjust $x$ according to our needs but he wasn't convinced. I hope there is a way to solve this question.
We just have to find out which options aren't possible for any value of $x$.
Above statement is my only progress(if the question is solved I indeed progressed further than my teacher XD).
Thanks for any hints or solution.
The only correct option is $(C)$.
$(A)$ is not correct since for $n=1,a=1,a_1=2,b=1,b_1=2$, there does not exist $x\in\mathbb R^+$ such that $$\log_x2-\log_x1=0$$
$(B)$ is not correct since for $n=1,a=2,a_1=1,b=1,b_1=2$, there does not exist $x\in\mathbb R^+$ such that $$\log_x1-\log_x2=0$$
For $(C)$, we see that $$\log_x a_n-\log_x a=b_n-b$$ is equivalent to $$\ln x=\frac{\ln\left(\frac{a_n}{a}\right)}{b_n-b}\tag1$$ where we always have $b_n-b=nd\not=0$ assuming that the common difference $d$ is not $0$.
Note here that the range of the function $y=\ln x$ is $\mathbb R$. So, for every real number $c$, there exists $x\in\mathbb R^+$ such that $\ln x=c$. So, there exists $x\in\mathbb R^+$ such that $(1)$. Hence, the option $(C)$ is correct.
$(D)$ is not correct since for $n=1,a=1,a_1=2,b=1,b_1=2$, there does not exist $x\in\mathbb R^+$ such that $$\log_x2-\log_x1=0$$