How do you solve $25^{1 - 3 n} = 625^{- n - 1}$ $25^ { 1- 3n } = 625^ { - n - 1}$ ?

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Answer 1 · 2017-09-22T05:05:53

$n = 3$ $n=3$

Rewrite the expression by finding the same base on both sides

$25^{1 - 3 n} = 625^{- n - 1}$ $25^(1-3n)=625^(-n-1)$

$\Rightarrow 5^{2 (1 - 3 n)} = 5^{4 (- n - 1)}$ $rArr5^(2(1-3n))=5^(4(-n-1))$

$\Rightarrow 5^{2 - 6 n} = 5^{- 4 n - 4}$ $rArr5^(2-6n)=5^(-4n-4)$

Equate the powers to solve for $n$ $n$

$2 - 6 n = - 4 n - 4$ $2-6n=-4n-4$

$2 n = 6$ $2n=6$

$n = 3$ $n=3$

How do you solve 251−3n=625−n−125^ { 1- 3n } = 625^ { - n - 1}?