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Mathematics 31 Online
OpenStudy (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

OpenStudy (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

OpenStudy (anonymous):

typo, the log product should read log ((X+4)(x-4))

OpenStudy (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 \[\Large \log _{9}(x^2-16)=1\]

OpenStudy (anonymous):

Yes, I got as far as @sami-21 But got stuck. I knew the answer was 5. Didnt know how to get it.

OpenStudy (anonymous):

so now we can cgange the log by change of bass formula

OpenStudy (anonymous):

change*

OpenStudy (anonymous):

Hmm..thats where i got stuck. How do you do that?

OpenStudy (anonymous):

\[\Large \log _{a}(b)=\frac{\ln(a)}{\ln{b}}\]

OpenStudy (anonymous):

it is upto use to change it to any base .i used natural log whose base is e

OpenStudy (anonymous):

Hmm...i think i get it.

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

I had another one similar to it: 7^(5x-2)=5^(3-x)

OpenStudy (anonymous):

I know you have to log both sides.

OpenStudy (anonymous):

yes take ln of both sides

OpenStudy (anonymous):

then use property \[\Large \log _{a}(b)^n=nlog _{a}b\] which means exponents gets down and gets multiplied !

OpenStudy (anonymous):

let me do the your question now using above property

OpenStudy (anonymous):

take ln of both sides\[\Large \ln(7)^{5x-2}=\ln(5)^{3-x}\] so using the above mentioned property \[\Large (5x-2)\ln(7)=(3-x)\ln(5)\]

OpenStudy (anonymous):

now use calculator to find values of ln7 and ln5

OpenStudy (anonymous):

ln7=1.946 ln5=1.609

OpenStudy (anonymous):

ln(7)=1.94 ln(5)=1.60 (5x-2)*(1.69)=(3-x)1.609

OpenStudy (anonymous):

can you solve this !!

OpenStudy (anonymous):

Yesss. I can take it from there, thank you so much! And the last one, so sorry, is b on the question.

OpenStudy (anonymous):

a typo ! correct is (5x-2)*(1.94)=(3-x)1.609

OpenStudy (anonymous):

ok. u mean this \[\Large \sqrt{3x-2}-\sqrt{x}=1\] ??

OpenStudy (anonymous):

Correct :)

OpenStudy (anonymous):

ok taking square of both sides

OpenStudy (anonymous):

Waaiit.

OpenStudy (anonymous):

we have \[\Large (\sqrt{3x-2}-\sqrt{x})^2=1^2\] using \[\Large (a-b)^2=a^2+b^2-2ab\] we have \[\Large ((3x-2)+x-2\sqrt{3x-2}\sqrt{x})=1\]

OpenStudy (anonymous):

Sorry, it was 3x-3

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Itsokay i will pretend the 2 was really 3

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Sorry :(

OpenStudy (anonymous):

its ok :) i am notorious for typo mistakes ask anyone here :P

OpenStudy (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

OpenStudy (anonymous):

Wow, thats good enough! Thank you soo much for your help. I learned a lot!

OpenStudy (anonymous):

Also, thank you too @GregTheo

OpenStudy (anonymous):

yw:)

OpenStudy (anonymous):

\[\sqrt(3x-2)-\sqrt(x)=1\] add \[\sqrt(x)\] in both sides

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