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Mathematics 8 Online
OpenStudy (anonymous):

i need help solving logarithmic equations

OpenStudy (anonymous):

\[\log_{5} 4-\log_{5} 3r=3\log_{5} 2\]

OpenStudy (anonymous):

you want to find r? ^_^

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

but i need step by step help so i can understand it.

OpenStudy (anonymous):

Alright, here we go :) :\[-\log_{5}3r = \log_{5}2^3 - \log_{5}4 \]\[\log_{5}3r = -\log_{5}2^3 +\log_{5}4 \]\[\log_{5}3r = \log_{5}(8.4) \]\[\log_{5}3r = \log_{5}32 \]\[\log_{5}3r = 2.15 \rightarrow 3r = 5^{2.15} \rightarrow r = \frac{5^{2.15}}{3} \approx10.6 \] correct me if I'm wrong please ^_^

OpenStudy (anonymous):

i'm not sure if it's right or wrong. that's my problem. i have no knowledge on this at all.

OpenStudy (anonymous):

hmm , alright the basic rules : \[\log_{b}n + \log_{b}k = \log_{b}(n + k) \] \[ \log_{b}n -\log_{b}k = \log_{b}(\frac{n}{k})\] I think I've done a silly mistake lol , hold on :)

OpenStudy (anonymous):

Use the properties of logerithms: \[\log _{5}4-\log _{5}3r=\log _{5}[4/(3r)]\] Also \[3\log _{5}2=3\log _{5}2^3=3\log _{5}8\] \[\log _{5} [4/(3r)]=\log _{5}8\] 4/3r=8 So r=1/6

OpenStudy (anonymous):

=// after the Also ignore the 3s after the equal signs

OpenStudy (anonymous):

this step :\[\log_{5} 3r = \log_{5}4 - \log_{5} 8\] must be : \[\log_{5}3r = \log_{5} \frac{4}{8}\]

OpenStudy (anonymous):

you can take it up from here, right? :) I think xav's answer is correct since I did a silly mistake lol, sorry ^_^"

OpenStudy (anonymous):

r=1/6

OpenStudy (anonymous):

Thats correct

OpenStudy (anonymous):

Thank you so much :)

OpenStudy (anonymous):

alright , so xav's + iam's answer = correct dear ^_^, did you understand the concept now?

OpenStudy (anonymous):

alright , so xav's + iam's answer = correct dear ^_^, did you understand the concept now?

OpenStudy (anonymous):

Well, you can verify if the answer is correct or not by going here http://www.wolframalpha.com/input/?i=log5%284%29%E2%88%92log5%283r%29%3D3log5%282%29

OpenStudy (anonymous):

Yes i did. I have plenty more problems i need help with. Can anyone be patient with me and help me through all of them ? :)

OpenStudy (anonymous):

what are your problems?

OpenStudy (anonymous):

What are your problems?

OpenStudy (anonymous):

\[\log_{2} 4-\log_{2} (x+3)-\log_{2} 8\]

OpenStudy (anonymous):

What do you want to do with this

OpenStudy (anonymous):

Simplify ?

OpenStudy (anonymous):

I guess solve for x

OpenStudy (anonymous):

Wait... >_<

OpenStudy (anonymous):

Where is the =

OpenStudy (anonymous):

No. all the problems i'm posting are solving each equation.

OpenStudy (anonymous):

Where is the =

OpenStudy (anonymous):

oh wait i see what you saying the = is supposed to be after (x+3)

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

\[2-log2(x+3)=3\]

OpenStudy (anonymous):

Where 2 is at the base

OpenStudy (anonymous):

yes 2 is the base.

OpenStudy (anonymous):

But that leads to no solution

OpenStudy (anonymous):

So instead if we use

OpenStudy (anonymous):

\[\log_{2} 4-\log_{2} 8=\log_{2} (x+3)\]

OpenStudy (anonymous):

We can write (4/8)=x+3

OpenStudy (anonymous):

The rest you can do

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