Please provide steps, having a bit of trouble.
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There are two 2^c terms in this equation so you could rewrite this like
2*(2^c) = 256
Since the bases are the same you can rewrite it under one exponent
2^(c+1) = 256
Now how many times can we multiply 2 to itself and yield 256 . . . 8 times. Hence
2^(c+1) = 2^8
The bases vanish by properties of exponents and you get the equation c+ 1 = 8. It must follow that c = 7 and you can test that
2^7 + 2^7 = 128 + 128 = 256
Yay!!
2*(2^c) = 256
Since the bases are the same you can rewrite it under one exponent
2^(c+1) = 256
Now how many times can we multiply 2 to itself and yield 256 . . . 8 times. Hence
2^(c+1) = 2^8
The bases vanish by properties of exponents and you get the equation c+ 1 = 8. It must follow that c = 7 and you can test that
2^7 + 2^7 = 128 + 128 = 256
Yay!!
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2^c + 2^c = 2 * 2^c = 2^1 * 2^c = 2^(c+1)
If 2^(c+1) = 256, take the ln of both sides
ln(2^(c+1) = ln(256)
(c + 1) ln (2) = ln (256)
c + 1 = ln(256) / ln (2)
c + 1 = 8 [Use a calculator if you want to compute the right hand side]
c = 7
(You don't have to use logs if you know 2^8 = 256, you can use 2^(c+1)=2^(8) and solve that way)
If 2^(c+1) = 256, take the ln of both sides
ln(2^(c+1) = ln(256)
(c + 1) ln (2) = ln (256)
c + 1 = ln(256) / ln (2)
c + 1 = 8 [Use a calculator if you want to compute the right hand side]
c = 7
(You don't have to use logs if you know 2^8 = 256, you can use 2^(c+1)=2^(8) and solve that way)
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since 2^c and 2^c are the same number, they are equal. So first you must divide 256 by two which is 128. So 2^c is equal to 128. Then you must find the exponent that makes 2 equal to 128.
So
128/2=64
64/2=32
32/2=16
16/2=8
8/2=4
4/2=2
2/2=1
so you have 7 times to divide by 2, so you must multiply 2 by two seven times to get 128. so c=7
So
128/2=64
64/2=32
32/2=16
16/2=8
8/2=4
4/2=2
2/2=1
so you have 7 times to divide by 2, so you must multiply 2 by two seven times to get 128. so c=7
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I'll answer in a way that's probably the most valuable to you, one that lets you work it out yourself without giving the exact answer...
Since the two terms on the left are the same they must each equal half of 256. How much is that?
Then what power of 2 do you need to reach that number, if 2^0 is 1, 2^1 is 2, 2^2 is 4, 2^3 is 8, etc?
Since the two terms on the left are the same they must each equal half of 256. How much is that?
Then what power of 2 do you need to reach that number, if 2^0 is 1, 2^1 is 2, 2^2 is 4, 2^3 is 8, etc?
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The above answerers are all wrong.
2^c + 2^c = 256
2^(c + 1) = 256
2^(c + 1) = 2^8
c + 1 = 8
x = 7
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I am receiving thumbs down for simply stating the original posters that posted their answer before me were incorrect. Guess people cannot stand criticism these days. Let's encourage and promote mistakes.
2^c + 2^c = 256
2^(c + 1) = 256
2^(c + 1) = 2^8
c + 1 = 8
x = 7
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I am receiving thumbs down for simply stating the original posters that posted their answer before me were incorrect. Guess people cannot stand criticism these days. Let's encourage and promote mistakes.
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2(2^c) = 256
2^(c+1) = 2^8
Therefore, c+1 = 8
c = 7
2^(c+1) = 2^8
Therefore, c+1 = 8
c = 7
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2 to the 7th power + 2 to the 7th power = 256.
2x2x2x2x2x2x2 = 128
128x2 = 256
Hope that helps
2x2x2x2x2x2x2 = 128
128x2 = 256
Hope that helps
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... 2^c + 2^c = 256
or 2* (2^c) = 256
or 2^c = 128
or 2^c = 2^7
or c = 7
or 2* (2^c) = 256
or 2^c = 128
or 2^c = 2^7
or c = 7
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... 2^c + 2^c = 2 x 2^c = 2^(1 + c) = 256
=> 2^(1 + c) = 2^8
=> 1 + c = 8
hence c = 7
=> 2^(1 + c) = 2^8
=> 1 + c = 8
hence c = 7
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8: sqrt(256)/2
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clog2 + clog2 = 256
c (log2 + log 2) = 256
c = 256/(log2 +log2)
c (log2 + log 2) = 256
c = 256/(log2 +log2)