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If 5^x+5^x+5^x+5^x+5^x=25^n, x=?

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If 5^x+5^x+5^x+5^x+5^x=25^n, x=?  [#permalink]

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New post 08 Jun 2017, 11:34
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(PS, algebra, exponent, 41~44)

If \(5^x+5^x+5^x+5^x+5^x=25^n\), x=?

A. n-1
B. n+1
C. 2-n
D. 2n
E. 2n-1

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Re: If 5^x+5^x+5^x+5^x+5^x=25^n, x=?  [#permalink]

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New post 08 Jun 2017, 11:44
\(5^x+5^x+5^x+5^x+5^x\), can be written as 5.\(5^x\) or\(5^{x+1}\)

Now it is given that \(5^x+5^x+5^x+5^x+5^x\) = \(25^n\) = \(5^{2n}\)

So

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

x+1 = 2n
x= 2n-1

Answer is E
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Re: If 5^x+5^x+5^x+5^x+5^x=25^n, x=?  [#permalink]

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New post 08 Jun 2017, 11:49
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Smokeybear00 wrote:
(PS, algebra, exponent, 41~44)

If \(5^x+5^x+5^x+5^x+5^x=25^n\), x=?

A. n-1
B. n+1
C. 2-n
D. 2n
E. 2n-1


Aside: some people won't recognize that the first step here it so collect like terms.
The reason for this is that we typically collect like terms when dealing with variables, and when we see this question, we don't see that we can apply the same rules.

For example, we know that k + k + k + k + k = 5k
And we know that y² + y² + y² + y² + y² = 5
And 3w + 3w + 3w + 3w + 3w = 5(3w) = 15w

Given: \(5^x+5^x+5^x+5^x+5^x=25^n\)

Apply the same rule to get: \(5(5^x) =25^n\)

Rewrite 5 as 5^1 to get: \((5^1)(5^x) =25^n\)

Combine exponents: \(5^{x+1} =25^n\)

Rewrite 25 as 5² to get: \(5^{x+1} =(5^2)^n\)

Apply power of a power rule: \(5^{x+1} =5^{2n}\)

Since the bases are equal, we can conclude that: \(x + 1 = 2n\)

Subtract 1 from both sides to get: \(x = 2n - 1\)

Answer:

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Re: If 5^x+5^x+5^x+5^x+5^x=25^n, x=?  [#permalink]

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New post 09 Jun 2017, 16:35
Smokeybear00 wrote:
(PS, algebra, exponent, 41~44)

If \(5^x+5^x+5^x+5^x+5^x=25^n\), x=?

A. n-1
B. n+1
C. 2-n
D. 2n
E. 2n-1


\(5^x+5^x+5^x+5^x+5^x\)
Taking \(5^x\) as common factor we get; \(5^x(1+1+1+1+1) = 5^x*5\)
Therefore; \(5^x*5 = 25^n\)
\(5^{x+1} = 5^{2n}\)
\(x + 1 = 2n\)
\(x = 2n - 1\). Answer E...
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Re: If 5^x+5^x+5^x+5^x+5^x=25^n, x=?  [#permalink]

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New post 09 Jun 2017, 22:50
Smokeybear00 wrote:
(PS, algebra, exponent, 41~44)

If \(5^x+5^x+5^x+5^x+5^x=25^n\), x=?

A. n-1
B. n+1
C. 2-n
D. 2n
E. 2n-1


\(5^x+5^x+5^x+5^x+5^x=25^n\)

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

\(5^x*5= 5^{2n}\)

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

So, \(x + 1 = 2n\)

Or, \(x = 2n - 1\)

Thus, answer will be (E) 2n-1
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If 5^x+5^x+5^x+5^x+5^x=25^n, x=?  [#permalink]

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New post 09 Jun 2017, 23:03
If \(5^x+5^x+5^x+5^x+5^x=25^n\)

On substituting x=1, we get 5 + 5 + 5 + 5 + 5= 25^n, so n=1
If we substitute x=2, we get 25+25+25+25+25 = 125 = 25^n.
When n=\(\frac{3}{2}\), \(25^\frac{3}{2}\) = \(\sqrt{25*25*25}\) = 125, so n=\(\frac{3}{2}\)

If we substitute x=3, we get 125+125+125+125+125 = 625 = 25^n, so n=2

This sequence of
x=1, n=1
x=2, n=3/2
x=3, n=2

translates to x = 2n -1 (Option E)
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Re: If 5^x+5^x+5^x+5^x+5^x=25^n, x=?  [#permalink]

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Re: If 5^x+5^x+5^x+5^x+5^x=25^n, x=? &nbs [#permalink] 27 Jul 2018, 00:19
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