i need help solving logarithmic equations

- anonymous

i need help solving logarithmic equations

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- anonymous

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

- anonymous

you want to find r? ^_^

- anonymous

yes.

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## More answers

- anonymous

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

- 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

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

- 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

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

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

- anonymous

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

- 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

r=1/6

- anonymous

Thats correct

- anonymous

Thank you so much :)

- anonymous

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

- anonymous

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

- 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

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

what are your problems?

- anonymous

What are your problems?

- anonymous

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

- anonymous

What do you want to do with this

- anonymous

Simplify ?

- anonymous

I guess solve for x

- anonymous

Wait... >_<

- anonymous

Where is the =

- anonymous

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

- anonymous

Where is the =

- anonymous

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

- anonymous

okay

- anonymous

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

- anonymous

Where 2 is at the base

- anonymous

yes 2 is the base.

- anonymous

But that leads to no solution

- anonymous

So instead if we use

- anonymous

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

- anonymous

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

- anonymous

The rest you can do

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