anonymous
  • anonymous
Have a few problems I was stuck on. 1)log_9(x+4)+log_9(x-4)=1 2)Sqrt(3x-2)-sqrt(x)=1
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
First combine the logs ( addition of log is multiplication) log(x+4) + log(x-4) = Log ((x+4)/(x-4)) raise everything from base 9 for part one, doing this cancels the log because they are inverse functions and 9^1 is just 9 resulting in (x+4)(x-4) = 9 distribute x^2 -16 = 9 x^2 = 25 x= + or -5
anonymous
  • anonymous
typo, the log product should read log ((X+4)(x-4))
anonymous
  • anonymous
ok lets solve it since \[\Large \log(a)+\log(b)=\log(ab)\] (addition changes to multiplication \[\Large \log _{9}((x+4)(x-4))=1\] since \[\Large a^2-b^2=(a+b)(a-b)\] so \[\Large \log _{9}(x^2-16)=1\]

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anonymous
  • anonymous
Yes, I got as far as @sami-21 But got stuck. I knew the answer was 5. Didnt know how to get it.
anonymous
  • anonymous
so now we can cgange the log by change of bass formula
anonymous
  • anonymous
change*
anonymous
  • anonymous
Hmm..thats where i got stuck. How do you do that?
anonymous
  • anonymous
\[\Large \log _{a}(b)=\frac{\ln(a)}{\ln{b}}\]
anonymous
  • anonymous
it is upto use to change it to any base .i used natural log whose base is e
anonymous
  • anonymous
Hmm...i think i get it.
anonymous
  • anonymous
ok
anonymous
  • anonymous
I had another one similar to it: 7^(5x-2)=5^(3-x)
anonymous
  • anonymous
I know you have to log both sides.
anonymous
  • anonymous
yes take ln of both sides
anonymous
  • anonymous
then use property \[\Large \log _{a}(b)^n=nlog _{a}b\] which means exponents gets down and gets multiplied !
anonymous
  • anonymous
let me do the your question now using above property
anonymous
  • anonymous
take ln of both sides\[\Large \ln(7)^{5x-2}=\ln(5)^{3-x}\] so using the above mentioned property \[\Large (5x-2)\ln(7)=(3-x)\ln(5)\]
anonymous
  • anonymous
now use calculator to find values of ln7 and ln5
anonymous
  • anonymous
ln7=1.946 ln5=1.609
anonymous
  • anonymous
ln(7)=1.94 ln(5)=1.60 (5x-2)*(1.69)=(3-x)1.609
anonymous
  • anonymous
can you solve this !!
anonymous
  • anonymous
Yesss. I can take it from there, thank you so much! And the last one, so sorry, is b on the question.
anonymous
  • anonymous
a typo ! correct is (5x-2)*(1.94)=(3-x)1.609
anonymous
  • anonymous
ok. u mean this \[\Large \sqrt{3x-2}-\sqrt{x}=1\] ??
anonymous
  • anonymous
Correct :)
anonymous
  • anonymous
ok taking square of both sides
anonymous
  • anonymous
Waaiit.
anonymous
  • anonymous
we have \[\Large (\sqrt{3x-2}-\sqrt{x})^2=1^2\] using \[\Large (a-b)^2=a^2+b^2-2ab\] we have \[\Large ((3x-2)+x-2\sqrt{3x-2}\sqrt{x})=1\]
anonymous
  • anonymous
Sorry, it was 3x-3
anonymous
  • anonymous
ok
anonymous
  • anonymous
Itsokay i will pretend the 2 was really 3
anonymous
  • anonymous
ok
anonymous
  • anonymous
Sorry :(
anonymous
  • anonymous
its ok :) i am notorious for typo mistakes ask anyone here :P
anonymous
  • anonymous
have a look at here http://www.wolframalpha.com/input/?i=solve%28sqrt%283x-3%29-sqrt%28x%29%3D1%29 its quite difficult to type :P
anonymous
  • anonymous
Wow, thats good enough! Thank you soo much for your help. I learned a lot!
anonymous
  • anonymous
Also, thank you too @GregTheo
anonymous
  • anonymous
yw:)
anonymous
  • anonymous
\[\sqrt(3x-2)-\sqrt(x)=1\] add \[\sqrt(x)\] in both sides

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