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- anonymous

Solve for x.
logx+log(x-15)=2

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- anonymous

Solve for x.
logx+log(x-15)=2

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- hartnn

hello :)
what did you try??
tried combining those 2 logs?

- anonymous

I know what you have to do, however, I am not sure how to do it :(

- anonymous

I don't know how to combine the logs

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- hartnn

\(\log A + \log B = \log AB \)
so how about log x + log (x-15)
?

- anonymous

is it 20?

- hartnn

you're asking if x = 20 ?
how did you do all the steps so fast! :O

- anonymous

I've been working on this problem for a really long time and I wasn't sure if I was doing the steps right so I posted on here. I think the answer is 20...I don't know what else it could be :(

- hartnn

oh great! you've tried some steps...
can you post them here, we'll correct then if required.

- anonymous

Oh boy... I don't know where to begin! I'll write the most recent steps I've tried...one moment

- anonymous

I used the formula \[a=\log _{b}(b ^{a})\]

- anonymous

\[2= \log _{10}(10^2)\]

- hartnn

for the right side?
so, \(2 = \log 10^2 \)

- hartnn

but for the left side we will need the formula I posted :)

- hartnn

\(\log x + \log(x-15) = \log x\times(x-15) \\= \log 10^2\)

- anonymous

\[\log _{10}(x)+\log _{10}(x-15)=\log _{10}(10^2)\]

- hartnn

good

- anonymous

\[\log _{10}(x(x-15))=\log _{10}(10^2)\]

- hartnn

you get how
\(\Large \log x + \log (x-15) = \log [x\times (x-15)]\)
basically just a log property

- anonymous

and the answeres are x=20 and x=-5 but we cant have a negative log so it must be 20

- hartnn

if log A = log B,
then A = B
what that says is that you can just cancel out log on both sides!
so we have
x(x-15) = 10^2
got this?

- anonymous

yes I just skipped the last steps at the end because they were really small and by then I already figured out my answers :)

- hartnn

oh you solved it completely.
excellent! x =20 is correct :)

- anonymous

Thanks @hartnn , I appreciate your time :)

- hartnn

welcome ^_^
happy to help :)

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