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):

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):

6 years ago
OpenStudy (anonymous):

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
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