If (3 + 4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^16 + 4^16)(3^32 + 4^32)

Question

If  (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 , what is the value of x?

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

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GMATinsight · Answer

For x = 2,  4^x - 3^x = 16 - 9 = 7  which is equal to  4+3 
For x = 4,  4^x - 3^x = 256 - 81 = 175  which is equal to  (4+3)(4^2+3^2)

OBSERVATION : Value of x is Double the power of 4 and 3 in biggest expression  (4^2+3^2)

i,e,  (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   i.e. x = 2*32 = 64

Answer: Option E

Lipun · Answer

(3 + 4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^16 + 4^16)(3^32 + 4^32)

= (4-3)(3+4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^16 + 4^16)(3^32 + 4^32)

=(4^2 - 3^2)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^16 + 4^16)(3^32 + 4^32)
.
. continue applying a^2 - b^2 = (a + b)(a - b)...after first two terms you can see that the successive term is twice in power
.
=4^64 - 3^64

So, x = 64

HoneyLemon · Answer

Lets find the value for the smaller terms in the sequence .

----- ----------- Term-------------------------------Value--------------------------X = ?-----------

------ 3+4--------------------------------------7 =  4 ^2-3^2  --------------------X = 2-----------

----(3+4)(3^2+4^2)-----------------------175 = 256-81= 4 ^4-3^4  ----------X = 4-----------

So we can apply this knowledge in for 
 (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 
 X must be 2 * 32 = 64 .

Ans is   E

chetan2u · Answer

Algebraic method 
The moment you see something of the form  a^2+b^2  and  a^x-b^x , try to use the formula =>  a^2-b^2=(a-b)(a+b) .

Now  a+b=\frac{a^2-b^2}{a-b}........a^2+b^2=\frac{a^4-b^4}{a^2-b^2}.......... a^4+b^4=\frac{a^8-b^8}{a^4-b^4}........a^{32}+b^{32}=\frac{a^{64}-b^{64}}{a^{32}-b^{32}}..........

(3 + 4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32})

=\frac{4^2-3^2}{4-3}*\frac{4^4-3^4}{4^2-3^2}*\frac{4^8-3^8}{4^4-3^4}*........*\frac{4^{64}-3^{64}}{4^{32}-3^{32}}

=\frac{4^{64}-3^{64}}{4-3}=4^{64}-3^{64}= 4^x - 3^x ..

Thus,  x=64

Choices 
Now each term in  (3 + 4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^{16} + 4^{16})(3^{32} + 4^{32})  is positive and will surely be greater than  3^{32} + 4^{32} 
and  3^{32} + 4^{32} > 4^{32} -3^{32} , so no choice  \leq{32}  can fit in.
 x=64

E

ORR1013 · Answer

Bunuel

I don't quite follow how we can just pull a 4-3 out of the air and tack it on to the left. Could use a bit of help understanding this problem as the answers make a few leaps that make me confused. Thanks in advance!

chetan2u · Answer

4-3 is nothing but 1, so We can write Anything that adds up to 1 to a product and it will not make any change. 
If we have x(x-7), we can write it as (7-6)(0.5+0.5)x(x-7).
Why? Because 7-6 is also 1 and 0.5+0.5 is also 1.

Here 4-3 is taken to relate it to 4+3 as (a-b)(a+b)=a^2-b^2

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If (3 + 4)(3^2 + 4^2)(3^4 + 4^4)(3^8 + 4^8)(3^16 + 4^16)(3^32 + 4^32)

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