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If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?

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Intern
Joined: 20 Jun 2019
Posts: 44
Location: United States (DC)
Schools: UVA Darden
GPA: 3.4
WE: Sales (Computer Software)
If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?  [#permalink]

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12 Mar 2020, 06:23
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If $$\frac{1}{5^x}$$ + $$\frac{20}{5^{x+1}}$$ = $$25^x$$, what is the value of x?

a) $$\frac{1}{2}$$

b)$$\frac{1}{3}$$

c)$$\frac{1}{4}$$

d)$$\frac{1}{5}$$

e)$$\frac{1}{6}$$
Intern
Joined: 20 Jun 2019
Posts: 44
Location: United States (DC)
Schools: UVA Darden
GPA: 3.4
WE: Sales (Computer Software)
Re: If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?  [#permalink]

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12 Mar 2020, 06:36
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I got the correct answer but not sure if I did the appropriate steps

$$\frac{1}{5^x}$$+$$\frac{20}{5^{x+1}}$$ = 25^x

$$\frac{1}{5^x}$$+$$\frac{5 * 2^2}{5^{x+1}}$$ =$$5^{2x}$$

$$\frac{1}{5^x}$$+$$\frac{5^x}{2^{-2}}$$ =$$5^{2x}$$

$$5^{-x}$$+$$\frac{5^x}{2^{-2}}$$=$$5^{2x}$$

pull out $$5^{-x}$$

$$5^{-x}$$ $$(1+\frac{-1}{2^{-2}})$$=$$5^{2x}$$

$$1+\frac{-1}{2^{-2}}$$ = 5

$$5^{-x}$$ * $$5^1$$ = $$5^{2x}$$

$$-x+1 = 2x$$

$$1 = 3x$$

$$x = \frac{1}{3}$$

ScottTargetTestPrep
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Joined: 02 Aug 2009
Posts: 8602
Re: If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?  [#permalink]

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12 Mar 2020, 06:48
whollymoses wrote:
If $$\frac{1}{5^x}$$ + $$\frac{20}{5^{x+1}}$$ = $$25^x$$, what is the value of x?

a) $$\frac{1}{2}$$

b)$$\frac{1}{3}$$

c)$$\frac{1}{4}$$

d)$$\frac{1}{5}$$

e)$$\frac{1}{6}$$

A quick way would be
$$\frac{1}{5^x}$$ + $$\frac{5*4}{5^{x+1}}$$ = $$25^x$$..
$$\frac{1}{5^x}$$ + $$\frac{4}{5^{x+1}*5^{-1}}$$ = $$25^x$$.....
$$\frac{1}{5^x}$$ + $$\frac{4}{5^{x}}$$ = $$25^x=5^{2x}$$.....
$$\frac{5}{5^x}$$ = $$5^{2x}$$.....
$$5=5^{2x+x}=5^{3x}$$
Equating the powers
$$1=3x...x=\frac{1}{3}$$
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If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?  [#permalink]

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12 Mar 2020, 06:51
whollymoses wrote:
If $$\frac{1}{5^x}$$ + $$\frac{20}{5^{x+1}}$$ = $$25^x$$, what is the value of x?

a) $$\frac{1}{2}$$

b)$$\frac{1}{3}$$

c)$$\frac{1}{4}$$

d)$$\frac{1}{5}$$

e)$$\frac{1}{6}$$

$$\frac{1}{5^x}+\frac{20}{5^x*5}=5^{2x}$$

$$\frac{5}{5^x}=5^{2x}$$

$$5^{3x}=5$$

$$3x=1$$

$$x=\frac{1}{3}$$

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Joined: 12 Dec 2015
Posts: 494
Re: If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?  [#permalink]

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12 Mar 2020, 07:55
If $$\frac{1}{5^x}$$ + $$\frac{20}{5^{x+1}}$$ = $$25^x$$, what is the value of x?

a) $$\frac{1}{2}$$

b)$$\frac{1}{3}$$ --> correct: $$\frac{1}{5^x}$$ + $$\frac{20}{5^{x+1}}$$ = $$25^x$$ => $$\frac{1}{5^x}$$ + $$\frac{20}{(5^x*5^1)}$$ = $$5^{2x}$$ => $$\frac{5}{5^x}$$ = $$5^{2x}$$ => 1-x = 2x => x=$$\frac{1}{3}$$

c)$$\frac{1}{4}$$

d)$$\frac{1}{5}$$

e)$$\frac{1}{6}$$
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Re: If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?  [#permalink]

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14 Mar 2020, 15:20
whollymoses wrote:
If $$\frac{1}{5^x}$$ + $$\frac{20}{5^{x+1}}$$ = $$25^x$$, what is the value of x?

a) $$\frac{1}{2}$$

b)$$\frac{1}{3}$$

c)$$\frac{1}{4}$$

d)$$\frac{1}{5}$$

e)$$\frac{1}{6}$$

Getting common denominators, we have:

5/(5^x * 5) + 20/(5^x * 5) = 5^(2x)

25/(5^x * 5) = 5^2x

5/5^x = 5^(2x)

5^(1 - x) = 5^2x

1 - x = 2x

1 = 3x

1/3 = x

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Re: If 1/5^x + 20/5^x+1 = 25^x, what is the value of x?   [#permalink] 14 Mar 2020, 15:20