Question

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

Answer 1

Is a) correct? and how do i do b?

Answer 2

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

Answer 3

thank you :D just didnt really know what "trial and error" really meant or wanted me to do

Answer 4

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.