Question

what is 2^log27?

Answer 1

if that's log base 2, then the log cancels the 2^ and the answer is 7

Answer 2

if this is
\[2^{\log(27)}\]it is a calculator exercise

Answer 3

if it is
\[2^{\log_2(7)}\] then the answer is 7

Answer 4

Use this formula
\[n^{\log_nm}=m\]

Answer 5

but i would hardly call composing a function with its inverse "canceling"

Answer 6

whta is log2(32)?

Answer 7

\[\log_b(b^x)=x\]
\[b^{\log_b(x)}=x\]because
\[f(f^{-1}(x))=x\]

Answer 8

\[\log_{2}32\]

Answer 9

rewrite 32 as 2^5, and the answer is 5

Answer 10

\[log_nn^m=m\]

Answer 11

if log10(3) approx. .4771,evaluate log10(.0003)

Answer 12

.0003 = 3*10^-4
log10 (3*10^-4)=log10(3) + log10(10^-4) = .4771 + -4 = -3.5229