Question

log81^(3-8)^k

Answer 1

can you draw this out?

Answer 2

so that -8 is not an exponent then?

Answer 3

sorry it is the exponent of the 3 in paranthesis

Answer 4

so it's really \[\Large \log_{81}\left(\left(3^{-8}\right)^k\right)\] ???

Answer 5

yes

Answer 6

\[ \Large \log_{81}\left(\left(3^{-8}\right)^k\right) \]


\[ \Large \log_{81}\left(3^{-8k}\right) \]


\[ \Large \frac{\log_{10}\left(3^{-8k}\right)}{\log_{10}\left(81\right)} \]


\[ \Large \frac{\log_{10}\left(3^{-8k}\right)}{\log_{10}\left(3^4\right)} \]


\[ \Large \frac{-8k\log_{10}\left(3\right)}{4\log_{10}\left(3\right)} \]


\[ \Large \frac{-8k\cancel{\log_{10}\left(3\right)}}{4\cancel{\log_{10}\left(3\right)}} \]


\[ \Large \frac{-8k}{4} \]


\[ \Large -2k \]


So \[ \Large \log_{81}\left(\left(3^{-8}\right)^k\right) = -2k \]

Answer 7

excellent
before i gave you good answer for this question, can you help me with this also
d) 10^(5log10)^(8−8log10)^5))

Answer 8

can you draw this out too? thx

Answer 9

jimthompson has a good answer. but if you have not gotten to change of base yet,
note that
\[3=81^{\frac{1}{4}}\]\]you can write
\[\log_{81}(3^{-8k})=\log_{81}(81^{-2k})=-2k\] because

Answer 10

So it's \[\Huge 10^{ 5\log_{10} \left( 8 \right) - 8\log_{10} \left( 5 \right) }\] ???