I have the question that if $\log_510=\log_7x(\log_nm)$ then values of $x$,$m$ and $n$ are?
This question looks easy but i tried to get the expression down to the form
$$\log_ab=\log_ac\tag{1.}$$
and i failed.
My attempt:-
the expression can be rewritten as :
$$\frac{\log_710}{\log_75}=\log_7{\frac{\log_7m^x}{\log_7n}}$$
After this step i don't get any clue to move forward.
Edit:-
The question has options one of which is the correct answer:-
(a) $10,7,5$
(b) $-1,2,3$
(c) $7,5,3$
(d) $7,5,8$
Help please....
HINT.- I hope this answer can help your inquire. If you take the function $$f(X)=\log_7(X)-\log_5 (10)$$ you have a unique and determined point, $a$, such that $f(a)=0$; this point is approximately equal to $16.183108$ so your “true” equation is $\log_n(m^x)=a$ really.
(Optionally you have $$f(X)=\log_7(X)-\frac{\log_7(10)}{\log_7(5)}= \log_7(10)-\log_7(10^{\frac{1}{\log_7(5)}}) =\log_7\left(\frac{X}{10^{\frac{1}{\log_7(5)}}}\right)=0$$ so you have $a=10^{\frac{1}{\log_7(5)}}\approx 16.183108$).
You have then to solve the equation $$\log_n(m^x)=a$$ If, for example (arbitrarily) $n=9$, the equation $\log_9(m^x)=a$ has a unique positive value, $b$, for $m^x$ since $a\gt 1$, $\log_9(x)$ is injective and its range is $\mathbb R$ so you have the equation $$m^x=b$$ Take now (again arbitrarily) $m=21$ and you have a unique $x$ such that $21^x=b$.
What conclusion can you have? See at Kamil 09875’s comment.