OpenStudy (anonymous):

Help please! If g(x)=log_7x, find g(1/49).

6 years ago
OpenStudy (anonymous):

what is the underscore supposed to represent?

6 years ago
OpenStudy (anonymous):

Just that 7 is lower.

6 years ago
OpenStudy (anonymous):

base 7 you mean?

6 years ago
OpenStudy (anonymous):

Yeah, my bad.

6 years ago
OpenStudy (anonymous):

g(1/49) = \[\log_{7} (1/49)\]

6 years ago
OpenStudy (anonymous):

uh no not quite, you can do it without one.

6 years ago
OpenStudy (anonymous):

\[\log_{7} (1) - \log_{7} (49)\]

6 years ago
OpenStudy (anonymous):

can you do the rest?

6 years ago
OpenStudy (anonymous):

Yea, thanks!

6 years ago
OpenStudy (anonymous):

True. Thanks

6 years ago
OpenStudy (anonymous):

yea but you can't always rely on a calculator. especially in university, you don't get to use one anyway.

6 years ago
OpenStudy (netlopes1):

if your function is \[g(x)=\log_{7}x\] and you want g(1/49) do this

6 years ago
OpenStudy (anonymous):

he already has the answer^ -.-.

6 years ago
OpenStudy (anonymous):

stop giving me notifications

6 years ago
OpenStudy (netlopes1):

\[g(1/49)=\log_{7}{1/49} \rightarrow \log_{7}{7^{-2}} \rightarrow -2.\log_{7}{7} \rightarrow \]

6 years ago
OpenStudy (netlopes1):

And finally, \[g(1/49)=-2.1 \rightarrow g(1/49) = -2\]

6 years ago
OpenStudy (anonymous):

Thanks! Wish I could give another medal.

6 years ago
OpenStudy (netlopes1):

Thanks!! If you need, i will for here, ok?

6 years ago