anonymous
  • anonymous
i need help solving logarithmic equations
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\log_{5} 4-\log_{5} 3r=3\log_{5} 2\]
anonymous
  • anonymous
you want to find r? ^_^
anonymous
  • anonymous
yes.

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anonymous
  • anonymous
but i need step by step help so i can understand it.
anonymous
  • 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 ^_^
anonymous
  • anonymous
i'm not sure if it's right or wrong. that's my problem. i have no knowledge on this at all.
anonymous
  • 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 :)
anonymous
  • 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
anonymous
  • anonymous
=// after the Also ignore the 3s after the equal signs
anonymous
  • anonymous
this step :\[\log_{5} 3r = \log_{5}4 - \log_{5} 8\] must be : \[\log_{5}3r = \log_{5} \frac{4}{8}\]
anonymous
  • anonymous
you can take it up from here, right? :) I think xav's answer is correct since I did a silly mistake lol, sorry ^_^"
anonymous
  • anonymous
r=1/6
anonymous
  • anonymous
Thats correct
anonymous
  • anonymous
Thank you so much :)
anonymous
  • anonymous
alright , so xav's + iam's answer = correct dear ^_^, did you understand the concept now?
anonymous
  • anonymous
alright , so xav's + iam's answer = correct dear ^_^, did you understand the concept now?
anonymous
  • 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
anonymous
  • 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 ? :)
anonymous
  • anonymous
what are your problems?
anonymous
  • anonymous
What are your problems?
anonymous
  • anonymous
\[\log_{2} 4-\log_{2} (x+3)-\log_{2} 8\]
anonymous
  • anonymous
What do you want to do with this
anonymous
  • anonymous
Simplify ?
anonymous
  • anonymous
I guess solve for x
anonymous
  • anonymous
Wait... >_<
anonymous
  • anonymous
Where is the =
anonymous
  • anonymous
No. all the problems i'm posting are solving each equation.
anonymous
  • anonymous
Where is the =
anonymous
  • anonymous
oh wait i see what you saying the = is supposed to be after (x+3)
anonymous
  • anonymous
okay
anonymous
  • anonymous
\[2-log2(x+3)=3\]
anonymous
  • anonymous
Where 2 is at the base
anonymous
  • anonymous
yes 2 is the base.
anonymous
  • anonymous
But that leads to no solution
anonymous
  • anonymous
So instead if we use
anonymous
  • anonymous
\[\log_{2} 4-\log_{2} 8=\log_{2} (x+3)\]
anonymous
  • anonymous
We can write (4/8)=x+3
anonymous
  • anonymous
The rest you can do

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