Question

logbase2 7
1.a rational number 2.an irrational number
3.a prime number 4.an integer

Answer 1

a

Answer 2

b

Answer 3

\[\frac{ \log7 }{ \log2 }=\frac{ 0.845 }{ 0.3010 }=2.80730879\]decide what it is

Answer 4

lets do it this way log base 2 (7)=x
2^x=7
it may be rational or irrational considering the long nature of above result
i think its irrational

Answer 5

It is 'b' but do you understand why?
logbase 2 of 7
means what power 'x' can you raise the number '2' to, and will equal 7?
so\[2^{x}=7\]
and you are solving for x
now if you list the first few of the sequence you get
\[2^{1}=2\]\[2^{2}=4\]\[2^{3}=8\]
So already 8 is bigger than 7

So you know it can't be an integer, which also means it can't be a prime
and I trust you are familiar with 2 to the power of a fraction?

Meaning it can only be 'irrational'

Answer 6

she could have guessed it from the final answer its b but still nice job sarah

Answer 7

Assume that \(\bf log_2(7)= \frac{m}{n}\) for positive integers m and n. Then:\[\bf 2^{\frac{ m }{ n }}=7 \implies 2^m = 7^n\]This is clearly a contradiction because the left side is always even and the right side is always odd. Not only that, but if we look at the graph of \(\bf 2^x\) and \(\bf 7^x\), we notice that they only ever intersect when x = 0 which implies that \(\bf m=n=0\) but then \(\bf \frac{m}{n}=\frac{0}{0}\) which is not possible hence \(\bf 2^m=7^n\) is a contradiction. This proves that \(\bf log_2(7)\) is irrational by contradiction.

Answer 8

@23ayesha1994

Answer 9

thanks chandan and sara

Answer 10

also genius

Answer 11

@sarahusher You're method of arriving at the answer being irrational is inconclusive but irrational is still the correct answer :).

Answer 12

@sarahusher i was taking the wrong ones out it was certainly not c or d hence i said so