GOODMAN
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
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GregTheo
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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
GregTheo
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typo, the log product should read log ((X+4)(x-4))
sami-21
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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
\[\Large \log _{9}(x^2-16)=1\]
GOODMAN
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Yes, I got as far as @sami-21
But got stuck. I knew the answer was 5. Didnt know how to get it.
sami-21
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so now we can cgange the log by change of bass formula
sami-21
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change*
GOODMAN
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Hmm..thats where i got stuck. How do you do that?
sami-21
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\[\Large \log _{a}(b)=\frac{\ln(a)}{\ln{b}}\]
sami-21
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it is upto use to change it to any base .i used natural log whose base is e
GOODMAN
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Hmm...i think i get it.
sami-21
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ok
GOODMAN
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I had another one similar to it:
7^(5x-2)=5^(3-x)
GOODMAN
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I know you have to log both sides.
sami-21
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yes take ln of both sides
sami-21
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then
use property
\[\Large \log _{a}(b)^n=nlog _{a}b\]
which means exponents gets down and gets multiplied !
sami-21
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let me do the your question now using above property
sami-21
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take ln of both sides\[\Large \ln(7)^{5x-2}=\ln(5)^{3-x}\]
so using the above mentioned property
\[\Large (5x-2)\ln(7)=(3-x)\ln(5)\]
sami-21
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now use calculator to find values of ln7 and ln5
GOODMAN
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ln7=1.946
ln5=1.609
sami-21
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ln(7)=1.94
ln(5)=1.60
(5x-2)*(1.69)=(3-x)1.609
sami-21
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can you solve this !!
GOODMAN
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Yesss. I can take it from there, thank you so much!
And the last one, so sorry, is b on the question.
sami-21
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a typo !
correct is
(5x-2)*(1.94)=(3-x)1.609
sami-21
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ok.
u mean this
\[\Large \sqrt{3x-2}-\sqrt{x}=1\]
??
GOODMAN
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Correct :)
sami-21
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ok
taking square of both sides
GOODMAN
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Waaiit.
sami-21
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we have
\[\Large (\sqrt{3x-2}-\sqrt{x})^2=1^2\]
using
\[\Large (a-b)^2=a^2+b^2-2ab\]
we have
\[\Large ((3x-2)+x-2\sqrt{3x-2}\sqrt{x})=1\]
GOODMAN
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Sorry, it was 3x-3
sami-21
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ok
GOODMAN
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Itsokay i will pretend the 2 was really 3
sami-21
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ok
GOODMAN
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Sorry :(
sami-21
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its ok :)
i am notorious for typo mistakes ask anyone here :P
GOODMAN
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Wow, thats good enough! Thank you soo much for your help. I learned a lot!
GOODMAN
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Also, thank you too @GregTheo
sami-21
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yw:)
muhammad9t5
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\[\sqrt(3x-2)-\sqrt(x)=1\]
add \[\sqrt(x)\] in both sides