Question

log7(1/7)

Answer 1

-1

Answer 2

because
\[7^{-1}=\frac{1}{7}\]

Answer 3

why would you put it to the -1? what about the log?

Answer 4

\[\log_b(x)=y\iff b^y=x\]

Answer 5

\[\log_7{(\frac{1}{7})} = \log_7({7^{-1}})=-1\log_7{(7)} = -1\]

Answer 6

you should be able to switch back and forth almost instantly. so you when you see
\[\log_7(\frac{1}{7})\] you are trying to fill in the question mark here
\[7^?=\frac{1}{7}\]

Answer 7

OKay thanks!

Answer 8

if you recognize
\[\frac{1}{7}\] as
\[7^{-1}\] and you can switch easily, then you can solve. if it is not clear that
\[7^{-1}=\frac{1}{7}\] or you can't go back and forth it will be a problem

Answer 9

True. Since the logarithm function 'reverses' exponentiation, returning the exponent of the base. In the case of log_7(1/7), it returns the value of the exponent x for a base 7 such that \[7^x = \frac{1}{7}\].

Answer 10

\[\log_{7}(1/7)=\log_{7}7^{-1} = -\log_{7} 7=-1(1)=-1 \]

Answer 11

log_n(whatever) returns a value x such that \[n^{x} = whatever\]

Answer 12

but n and whatever must be > 0