Question #7d0d9

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Question #7d0d9

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Answer 1 · 2017-05-27T15:12:01

$n = 1$ $n = 1$

Explanation:

I'm going to assume the question is $5^{n + 1} = \sqrt{25^{3 n - 1}}$ $5^(n + 1) = sqrt(25^(3n - 1))$

Square both sides to get:

${(5^{n + 1})}^{2} = {(\sqrt{25^{3 n - 1}})}^{2}$ $(5^(n + 1))^2 =(sqrt(25^(3n - 1)))^2$

Use ${(a^{n})}^{m} = a^{n m}$ $(a^n)^m = a^(nm)$

$5^{2 n + 2} = 25^{3 n - 1}$ $5^(2n + 2) = 25^(3n - 1)$

$5^{2 n + 2} = {(5^{2})}^{3 n - 1}$ $5^(2n + 2) = (5^2)^(3n - 1)$

$5^{2 n + 2} = 5^{6 n - 2}$ $5^(2n + 2) = 5^(6n - 2)$

$2 n + 2 = 6 n - 2$ $2n + 2 = 6n - 2$

$4 = 4 n$ $4 = 4n$

$1 = n$ $1= n$

Hopefully this helps!

Answer 2 · 2017-05-27T15:56:09

$n = 1$ $color(brown)(n=1$

Explanation:

$5^{n + 1} = {(\sqrt{25})}^{3 n - 1}$ $5^(n+1)=(sqrt25)^(3n-1)$

$:.5^(n+1)=5^(3n-1)$

base numbers are the same then:

$:.n+1=3n-1$

$:.n-3n=-1-1$

$:.-2n=-2$

multiply both sides by $-1$

$:.2n=2$

$:.color(brown)(n=1$

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