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# If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is value x/y?

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If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is value x/y?  [#permalink]

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Updated on: 25 Oct 2017, 05:48
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25% (medium)

Question Stats:

82% (02:26) correct 18% (02:56) wrong based on 169 sessions

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If $$(4^x)(8^y)(16^{7x}) = (8^{4y})(32^x)(64^y)$$ what is the value of x/y?

A. 3/5
B. 2/3
C. 3/4
D. 1
E. 4/3

Hi,
I want to know whether there is quicker solution for this question, please.

[With standard math]
As (b^c)^d = b^cd, and (b^c)*(b^d) = b^(c+d)
(2^2x)(2^3y)((2^4)^7x) = ((2^3)4y)(2^5x)(2^6y) → (2^2x)(2^3y)(2^28x) = (2^12y)(2^5x)(2^6y)
(2x + 3y + 28x) = (12y + 5x + 6y) → (30x + 3y) = (18y + 5x)
(25x = 15y) → (5x = 3y)
(x/y) = (3/5)

Originally posted by goodyear2013 on 05 Jan 2014, 07:39.
Last edited by Bunuel on 25 Oct 2017, 05:48, edited 3 times in total.
Edited the question.
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Joined: 02 Sep 2009
Posts: 50058
Re: If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is value x/y?  [#permalink]

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05 Jan 2014, 07:48
1
1
goodyear2013 wrote:
If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is the value of x/y?

A. 3/5
B. 2/3
C. 3/4
D. 1
E. 4/3

Hi,
I want to know whether there is quicker solution for this question, please.

[With standard math]
As (b^c)^d = b^cd, and (b^c)*(b^d) = b^(c+d)
(2^2x)(2^3y)((2^4)^7x) = ((2^3)4y)(2^5x)(2^6y) → (2^2x)(2^3y)(2^28x) = (2^12y)(2^5x)(2^6y)
(2x + 3y + 28x) = (12y + 5x + 6y) → (30x + 3y) = (18y + 5x)
(25x = 15y) → (5x = 3y)
(x/y) = (3/5)

$$(4^x)(8^y)(16^{7x}) = (8^{4y})(32^x)(64^y)$$;

$$(2^{2x})(2^{3y})(2^{28x}) = (2^{12y})(2^{5x})(2^{6y})$$;

$$2^{2x+3y+28x} = 2^{12y+5x+6y}$$;

$$2x+3y+28x=12y+5x+6y$$;

$$25x=15y$$;

$$\frac{x}{y}=\frac{3}{5}$$.

Theory on Exponents: math-number-theory-88376.html

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Tough and tricky PS exponents and roots questions with detailed solutions: tough-and-tricky-exponents-and-roots-questions-125956.html

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Re: If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is value x/y?  [#permalink]

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25 Feb 2014, 23:27
LHS >> 4, 8, 16 all are powers of 2 re-written as 2^2, 2^3 & 2^4 respectively

RHS >> 8,32, 64 all are powers of 2, re-written as 2^3, 2^5 & 2^6 respectively

After making bases same, powers can be equated:

2x + 3y + 28x = 12y + 5x + 6y

25x = 15y

x/y = 3/5 = Answer = A
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Re: If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is value x/y?  [#permalink]

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29 Oct 2017, 07:19
goodyear2013 wrote:
If $$(4^x)(8^y)(16^{7x}) = (8^{4y})(32^x)(64^y)$$ what is the value of x/y?

A. 3/5
B. 2/3
C. 3/4
D. 1
E. 4/3

Let’s simplify the given equation by expressing all the bases in terms of powers of 2:

(2^2x)(2^3y)(2^28x) = (2^12y)(2^5x)(2^6y)

2^(30x + 3y) = 2^(18y + 5x)

30x + 3y = 18y + 5x

25x = 15y

x/y = 15/25 = 3/5

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Re: If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is value x/y?  [#permalink]

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29 Oct 2017, 07:42
goodyear2013 wrote:
If $$(4^x)(8^y)(16^{7x}) = (8^{4y})(32^x)(64^y)$$ what is the value of x/y?

A. 3/5
B. 2/3
C. 3/4
D. 1
E. 4/3

$$(4^x)(8^y)(16^{7x}) = (8^{4y})(32^x)(64^y)$$

Or, $$(2^{2x})(2^{3y})(2^{28x}) = (2^{12y})(2^{5x})(2^{6y})$$

Or, $$2^{30x + 3y} = 2^{5x + 18y}$$

Or, $$25x = 15y$$

Or, $$5x = 3y$$

So, $$\frac{x}{y}= \frac{3}{5}$$

Thus, the correct answer will be (A)...

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Re: If (4^x)(8^y)(16^7x) = (8^4y)(32^x)(64^y) what is value x/y? &nbs [#permalink] 29 Oct 2017, 07:42
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