Question

????

Answer 1

\[\log_{5}16 + \log_{5}4 - 5\log_{5}2 - \log_{5}2 \]

Answer 2

simplify

Answer 3

how would you approach this @Yttrium

Answer 4

Do you remember the identity \[\log_{a}b = \frac{ logb }{ loga } \]
Just apply this first. And let's see where will we arrive.

Answer 5

\[\frac{ \log_{16} }{ \log_{5} } + \frac{ \log_{4} }{ \log_{5} } - \log_{5}2^{5} - \frac{ \log_{2} }{ \log_{5} }\]
\[\frac{ \log_{16} }{ \log_{5} } + \frac{ \log_{4} }{ \log_{5} } - \frac{ \log_{32} }{ \log_{5} } - \frac{ \log_{2} }{ \log_{5} }\]

Answer 6

i think i know what happens next. let me see

Answer 7

Sure. :))

Answer 8

any guidance @Yttrium i'm not sure what to do next, what i had in mind is not gonna work

Answer 9

Did you see that all of them in the numerator can be simplified as log of 2? :)

Answer 10

no, kidding. how is it that one miss small things like that? lol

Answer 11

\[\frac{ 2\left( \log_{8} +\log_{2} + \log_{16} - \log_{1} \right) }{ \log_{5} } \]

something like this perhaps?

Answer 12

\[\boxed{\log_b x+\log_by=\log_bxy}\]\[\boxed{\log_b x-\log_by=\log_b\dfrac xy}\]

Answer 13

\[\frac{ \log_{^{16\times4}} - \log_{\frac{ 32 }{ 2 }} }{ \log_{5} }\]

Answer 14

\[\frac{ \log_{64} - \log_{16} }{ \log_{5} }\]

Answer 15

\[\frac{ \log_{\frac{ 64 }{ 16}} }{ \log_{5} }\]

Answer 16

\[\frac{ \log_{4} }{ \log_{5} }\]

Answer 17

is this in its simplest form ?

Answer 18

is this even correct? i have a feeling im way off