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Solve log2 16 – x = 7
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Is it 16 or is it (16-x)?
16-x
Solve log2 16 – x = 7 log2 (16 - x) = log2 (128) 16-x=128 x = -112 x=-3
Okay...then it's \[2^{7}=16-x\]
128 = 16 - x x = -112
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\[\log_{2}(16-x) =7\]?? Right?
I am not understanding how the answer can b -3
Yes
\[\log_{2}16-x=7 \]
1st one or 2nd one?
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2nd
If it's the 2nd (more likely) option then yes, it is -3.
x = -112
Corey how are you getting -3
Here are the steps:
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I didn't solve for x at the end because it's simple algebra...but... x + 7 = 4 -7 -7 ____________ x = -3
Hope the image helped somewhat.
Fantastic! Thank you
Not a problem!
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