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Find the value of x

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Find the value of x  [#permalink]

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New post 05 Nov 2018, 22:30
4
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

70% (01:36) correct 30% (01:47) wrong based on 91 sessions

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Find the value of x such that

\((3+4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

A. 4
B. 8
C. 16
D. 32
E. 64

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Re: Find the value of x  [#permalink]

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New post 05 Nov 2018, 23:09
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eswarchethu135 wrote:
Find the value of x such that

\((3+4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

A. 4
B. 8
C. 16
D. 32
E. 64


Recognizing the pattern, is best way to solve this type of problems

Here pattern is \((a-b)*(a+b) = (a^2 - b^2)\)

Therefore, as (4-3) = 1, multiplying L.H.S by 1,we have
\((3+4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)
=> \((1)(3+4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

=> \((4-3)(3+4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

=> \((4^2 - 3^2)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

=> \((4^4 - 3^4)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

=> \((4^8 - 3^8)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

=> \((4^{16} - 3^{16})(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

=> \((4^{32} - 3^{32})(3^{32} + 4^{32}) = 4^x - 3^x\)

=> \((4^{64} - 3^{64}) = 4^x - 3^x\)

=> x = 64

Answer choice E

Kudos, if you like explanation

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Re: Find the value of x  [#permalink]

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New post 05 Nov 2018, 22:44
eswarchethu135 wrote:
Find the value of x such that

\((3+4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32}) = 4^x - 3^x\)

A. 4
B. 8
C. 16
D. 32
E. 64

Sol: I tried this question by putting values and proving RHS =LHS

I tries from largest value available in the answer choice
let x=64
then RHS= 4^64-3^64
= (4^32)^2-(3^32)^2
=(4^32-3^32) (4^32+3^32)
=(4^16-3^16) (4^16+3^16) (4^32+3^32)
=(4^8-3^8) (4^8+3^8)(4^16+3^16) (4^32+3^32)
=(4^4-3^4)(4^4+3^4) (4^8+3^8)(4^16+3^16) (4^32+3^32)
=(4^2-3^2)(4^2+3^2)(4^4+3^4) (4^8+3^8)(4^16+3^16) (4^32+3^32)
=(4-3)(4+3)(4^2+3^2)(4^4+3^4) (4^8+3^8)(4^16+3^16) (4^32+3^32)
=(4+3)(4^2+3^2)(4^4+3^4) (4^8+3^8)(4^16+3^16) (4^32+3^32)
= LHS
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Re: Find the value of x  [#permalink]

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New post 15 Jan 2019, 19:47
Hello !

Could someone please help me with this question?

Kind regards!
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Find the value of x  [#permalink]

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New post 15 Jan 2019, 20:06
2
jfranciscocuencag wrote:
Hello !

Could someone please help me with this question?

Kind regards!


Hi jfranciscocuencag

You just have to remember the basic formula that \((a^2 - b^2)\) = (a-b) (a+b) ------------(a)

Now you need not even solve the whole expression of \(4^x - 3^x\)to get to the LHS

Since LHS has the power \(3^{32} + 4^{32}\), this means that value of x would have been 64

\((4^{64} - 3^{64})\) = \((4^{32} - 3^{32})\) \((4^{32} + 3^{32})\),

Each time when you will expand the above expression using (a), you will be diving the power by half(1/2).

I hope this addresses your doubt
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Re: Find the value of x  [#permalink]

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New post 15 Jan 2019, 20:15
1
KanishkM wrote:
jfranciscocuencag wrote:
Hello !

Could someone please help me with this question?

Kind regards!


Hi jfranciscocuencag

You just have to remember the basic formula that \((a^2 - b^2)\) = (a-b) (a+b) ------------(a)

Now you need not even solve the whole expression of \(4^x - 3^x\)to get to the LHS

Since LHS has the power \(3^{32} + 4^{32}\), this means that value of x would have been 64

\((4^{64} - 3^{64})\) = \((4^{32} - 3^{32})\) \((4^{32} + 3^{32})\),

Each time when you will expand the above expression using (a), you will be diving the power by half(1/2).

I hope this addresses your doubt


Now is so clear KanishkM , thank you! +1KUD
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Re: Find the value of x   [#permalink] 15 Jan 2019, 20:15
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