Question

express as a sum of logarithms

log v4 (38*19)

Answer 1

hint \[\huge \log_2 (3 \times 4) \implies \log_2 3 + \log_2 4\]
does that help?

Answer 2

sadly no. i dont understand log that well in general and have no idea how to plug it into a calculator

Answer 3

i have a ti-83 plus that i got from a friend, but im not too familiar with graphing calculators

Answer 4

you don't need to use calculators here. just laws of logarithms

Answer 5

\[\huge \log_a (bc) = \log_a b + \log_a c\]
therefore...
\[\huge \log_ 2 (3 \times 4) = \log_2 3 + \log_2 4\]
making sense now?

Answer 6

\[\huge \log_3 ( 2\times 4) = \log_3 2 + \log_ 3 4\]

Answer 7

so now.. can you find \[\huge \log_4 (38 \times 19)\]

Answer 8

yeah so i ended up with \[\log_{4}38+\log_{4} 19\] what next?

Answer 9

that's it. that's the answer

Answer 10

ok. thanks once again.

Answer 11

welcome

Answer 12

what about \[\ln \frac{ 8 }{ 9x ^{9}y}\]

Answer 13

does it work the same way?

Answer 14

what's the instruction?

Answer 15

express in terms of sums and differences of logarithms

Answer 16

sum works same way...but difference no \[\huge \ln (\frac ab) \implies \ln a - \ln b\]
NOTE:
\[\LARGE \ln (\frac a{bc}) \implies \ln a - (\ln b + \ln c) \implies \ln a - \ln b - \ln c\]
do you get what i mean?

Answer 17

k. i think i get it. thanks again

Answer 18

welcome

Answer 19

ugh.....everytime i think it gets easier it just gets harder.....can you help me with this too. I feel like i should be paying you for this.

Express as a sum or difference of logarithms without exponents
\[\log _{c5}\sqrt{\frac{ x ^{2} }{ y ^{5}z ^{7}}}\]

Answer 20

without exponents...so this time you're going to use power rule...

Answer 21

i think i should just give a similar example..it's hard to explain in words

Answer 22

\[\large \ln \sqrt{\frac{a^3}{b^2c}} \implies \ln (\frac{a^3}{b^2 c})^{\frac 12} \implies \frac 12 \ln (\frac{a^3}{b^2 c}) \implies \frac 12 \ln a^3 - \frac 12 \ln b^2 - \frac 12 \ln c\]
\[\large \implies \frac 32 \ln a- \frac 22 \ln b - \frac 12 ln c \implies \frac 32 \ln a - \ln b - \frac 12 \ln c\]

Answer 23

the power rule is this: \[\huge \log_a b^c \implies c\log_a b\]