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Bunuel
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

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
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Bunuel
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


Are You Up For the Challenge: 700 Level Questions

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.
\(x=64\)

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Lipun
(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

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!
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Lipun
(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

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!

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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