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Re: If n is an integer,and 5n+5n=62625,how many values of n satisfy the e [#permalink]
5^n + 1/5^n = (625+1)/25
5^n +1/5^n = 625/25 + 1/25
5^n +1/5^n = 25 + 1/25
5^n +1/5^n = (5)^2 + 1/5^2

Now, N=2
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If n is an integer,and 5n+5n=62625,how many values of n satisfy the e [#permalink]
jai3892 wrote:
5^n + 1/5^n = (625+1)/25
5^n +1/5^n = 625/25 + 1/25
5^n +1/5^n = 25 + 1/25
5^n +1/5^n = (5)^2 + 1/5^2

Now, N=2

I think this is wrong approach. Question stem didn't ask for the value of N, rather it asked for how many values of N satisfy the equation.
According to your result, the answer would be 1. Since you got N=2 only.

It can be modified from the last step, if I am not mistaken.

5^n+5^-n= 5^2 + 5^-2 , thus n has two values (2 and -2)
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If n is an integer,and 5n+5n=62625,how many values of n satisfy the e [#permalink]
jonNNNNNy wrote:
anyone with another approach plz?




\(5^n + 5^{-n}=\frac{625}{25}\)

\(5^n +\frac{ 1}{5^n}=\frac{5^4}{5^2}\)

\(\frac{5^{2n}+1}{5^n}=\frac{5^4}{5^2}\)

cross multiplication \(5^2(5^{2n}+1)=5^4*5^n\)

\(5^{4n}+5^2=5^n*5^4\)

\(\frac{5^{4n}}{5^n*5^4}=-5^2\)

\(\frac{5^{3n}}{5^4}=-5^2\)

\(5^{3n}=5^4*25\)

\(5^{3n}=5^6\)

so we can eliminate the bases having

3n=6

n=2
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If n is an integer,and 5n+5n=62625,how many values of n satisfy the e [#permalink]
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