Write 2log35 + log32 as a single logarithm
\[2\log _{3}5 + \log _{3}2\]
will give medal
logarithmic properties, coocoobird
alright could you give me the property i need to use for this equation
i don't know what to do with the 2 in front of the first log
\[\Large b\cdot\log(a) \quad=\quad \log(a^b)\]
it looks like this property logb(xy) = logbx + logby. but what happens to the 2
b=2
well, you have to read the contents of the link I gave you
then you would know what to do
is 3 a?
no. 3 is the base. The rule i posted above deals with the 2 and 5.
\[alog_{n} r = \log_{n} r^{a} \] and \[\log_{n} a + \log_{n} b = \log_{n} (ab) \]. @katherinesmith these are the two properties you need for this particular question.
there's nothing that they're telling you now that is not in the link…
so i'm down to 3 * log(2) = log8
now what
read
3log(2)=log(8)?? :O
lol
@nincompoop if i wanted a link to a website about logarithms i could've googled one. i'm on this website because i need serious help with math because logarithms are not easy to understand when you have to learn about them online. so stop telling me to read.
2^3 = 8 dohhhhhhhhhhhhhhh
Katherine, i've been posting the same formulas several times on your posts. You just don't seem to be getting it. It would really be a good idea to read up on logs maybe :c
they're easy to understand, but you need to read about them first. this website is not for spoon-feeding solutions
i don't want spoon feeding, i understand the process of having formulas you have to go through but i don't know which ones to use. i don't know.
@nincompoop the link is too general for this situation. Give her something particularly suited for this situation.
it's the same in all cases
@katherinesmith did you see the exact formulas you need for this problem?
Example of the rule I posted above:\[\Large 3\cdot \log_4(x) \quad=\quad \log_4(x^3)\] Will that help you apply it to your own problem maybe kath? :o See which numbers are going where?
yes thank you for helping me.
i'm gonna work on it and close the question. i don't need tension over a math problem. i have plenty other questions to ask.
Join our real-time social learning platform and learn together with your friends!