Find $n$ if $7^{3n-1}-343^{\,n-1}=2352$.
Answer and solution
Correct answer: D
Solution
$343^{n-1}=7^{3(n-1)}=7^{3n-3}$.
$7^{3n-1}-7^{3n-3}=7^{3n-3}(7^2-1)=7^{3n-3}\times48=2352$.
$7^{3n-3}=49=7^2 \Rightarrow 3n-3=2 \Rightarrow n=\dfrac{5}{3}$.
$7^{3n-1}-7^{3n-3}=7^{3n-3}(7^2-1)=7^{3n-3}\times48=2352$.
$7^{3n-3}=49=7^2 \Rightarrow 3n-3=2 \Rightarrow n=\dfrac{5}{3}$.