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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
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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Nick88888888
 one year ago
Best ResponseYou've already chosen the best response.1Is a) correct? and how do i do b?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1you 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

Nick88888888
 one year ago
Best ResponseYou've already chosen the best response.1thank you :D just didnt really know what "trial and error" really meant or wanted me to do

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I 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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