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Bunuel
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We have to take common as we can't do anything to exponents separated via addition sign

Therefore,
2^199(1+1) = 2^199*2

Now we can add them, as 2= 2^1

So, 2^(199+1)
2^200=2^x
X= 200 (B)

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2^199+2^199=2^x
Let y=2^199
y+y=2^x
2y=2^x
2*2^199=2^x
2^200=2^x
Therefore x=200
Hence B.

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Bunuel
If \(2^{199} + 2^{199} = 2^x\), what is the value of x?

A. 199
B. 200
C. 201
D. 202
E. 203

2^199 is a common factor of both terms on the left side of the equation. Simplifying, we have:

2^199(1 + 1) = 2^x

2^199 * 2^1 = 2^x

2^200 = 2^x

200 = x

Answer: B
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