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Nick88888888

  • one year ago

A) Convert the following expression to exponential form: log2 128 = k B) Using trial and error, find the correct value of k. Prove that your answer is correct. Ive done this so far A) if y= b^x, then x=logb y Log2 128= k ^Base ^exponent 2^k=128 B) 2^7=128

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  1. Nick88888888
    • one year ago
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    Is a) correct? and how do i do b?

  2. UsukiDoll
    • one year ago
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    you have the answer already. to convert \[\log_2(128)=k\] into exponential form we need \[b^e=n\] (where b is the base, e is the exponent and n is the number) so base is 2 the number is 128 and the exponent is k \[2^k =128 \] then we figure out what k needs to be to obtain 128 so it's a high exponential number \[2^7 = 128 \] is correct. no trial and error stuff and to prove that just write 2 7 times 2 x 2 x 2 x 2 x 2 x 2 x 2 =128 128=128

  3. Nick88888888
    • one year ago
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    thank you :D just didnt really know what "trial and error" really meant or wanted me to do

  4. UsukiDoll
    • one year ago
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    I think it meant to let k be a smaller number and see what happens or something like that but we already found k to be 7, so it wasn't needed.

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