Question

What is the value of $n$ if ${343}^{\frac{n}{3}}\times 7^{5+3n}\times 7^{-2n}\times {49}^{n-1}=343$ ?

• A. 0
• B. 3
• C. -3
• D. 2

Answer 1

The correct option is A

0

${343}^{\frac{n}{3}}\times 7^{5+3n}\times 7^{-2n}\times {49}^{n-1}=343$

$(7^3{)}^{\frac{n}{3}}\times 7^{5+3n}\times 7^{-2n}\times (7^2{)}^{n-1}=7^3$

$7^n\times 7^{5+3n}\times 7^{-2n}\times 7^{2n-2}=7^3$

$7^{n+5+3n-2n+2n-2}=7^3$

$7^{4n+3}=7^3$ [The bases are equal. Therefore, the powers will be equal] ​

$4n+3=3$

$4n=0$

$n=0$