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Find the value of x

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Manager
Joined: 13 Jan 2018
Posts: 241
Location: India
Concentration: Operations, General Management
GMAT 1: 580 Q47 V23
GMAT 2: 640 Q49 V27
GPA: 4
WE: Consulting (Consulting)
Find the value of x  [#permalink]

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05 Nov 2018, 22:30
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Difficulty:

25% (medium)

Question Stats:

71% (01:32) correct 29% (01:42) wrong based on 77 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

_________________

____________________________
Regards,

Chaitanya

+1 Kudos

if you like my explanation!!!

Manager
Joined: 01 Mar 2015
Posts: 75
Re: Find the value of x  [#permalink]

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05 Nov 2018, 23:09
6
1
1
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

Kudos, if you like explanation

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General Discussion
Manager
Joined: 09 Jul 2018
Posts: 65
Re: Find the value of x  [#permalink]

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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
Manager
Joined: 12 Sep 2017
Posts: 175
Re: Find the value of x  [#permalink]

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15 Jan 2019, 19:47
Hello !

Kind regards!
VP
Joined: 09 Mar 2018
Posts: 1007
Location: India
Find the value of x  [#permalink]

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

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).

_________________

If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
Many of life's failures happen with people who do not realize how close they were to success when they gave up.

Manager
Joined: 12 Sep 2017
Posts: 175
Re: Find the value of x  [#permalink]

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

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).

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