anonymous
  • anonymous
solve for x: log(5)(11-6x) = log(5)(1-x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Nnesha
  • Nnesha
hint: \[\huge\rm log_b x = \log_b y\] \[\huge\rm\cancel { log_b} x = \cancel{\log_b} y\] x=y
anonymous
  • anonymous
Ohhh... X = 2 right?
anonymous
  • anonymous
Or is it no solution?

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Pawanyadav
  • Pawanyadav
Is 5 it's base
Nnesha
  • Nnesha
sorry i forgot :(
anonymous
  • anonymous
Yes 5 is the base
anonymous
  • anonymous
That's ok Nnesha. Thanks for trying to help :)
Nnesha
  • Nnesha
no i'm saying i forgot i ws helping u :P
anonymous
  • anonymous
Oh hahaha
Nnesha
  • Nnesha
alright so what was your first step ?? :-)
anonymous
  • anonymous
I would cross out the bases because they are the same right?
Nnesha
  • Nnesha
yep right and then simple algebra!!! :-)
Pawanyadav
  • Pawanyadav
16-6x=1-x 15=5x x=?
Nnesha
  • Nnesha
and yes you got it right :-)
mathmate
  • mathmate
Hint: dom log(x) = (0,inf)
anonymous
  • anonymous
Oh this is a tricky no solution problem isn't it....
mathmate
  • mathmate
... because?
anonymous
  • anonymous
The base is 5 instead of 10?
mathmate
  • mathmate
not exactly. Use the hint!
anonymous
  • anonymous
Oh! I would have no 'y' value right?
Nnesha
  • Nnesha
nope when there is just log then u have to suppose 10 base
mathmate
  • mathmate
actually, the hint applies to any base, so it was not indicated.
anonymous
  • anonymous
Oh ok, so the answer would just be x = 2?
mathmate
  • mathmate
want further hint? @lehmad
anonymous
  • anonymous
yes please :p
Nnesha
  • Nnesha
i'm confused do you have to solve for x right ?
mathmate
  • mathmate
Hint: Whenever you solve an equation involving log, you need to substitute back into the equation to reject all roots that make log negative (recall dom log(x)=(0,inf) ).
mathmate
  • mathmate
@Nnesha yes, the solution x=2 is correct for the algebraic part, but not correct for the problem given!
anonymous
  • anonymous
I have to solve for the problem given @Nnesha
anonymous
  • anonymous
Oh ok, so when I plug x=2 into the final product, I get f(2) = (0,inf) which is not solvable so there is no solution
mathmate
  • mathmate
\(log_5(11-6x)=log_5(-1)\) so x=2 is not an admissible solution because log(-1) is outside the domain of log.
mathmate
  • mathmate
* -1 is outside the domain of log.
mathmate
  • mathmate
sorry, gtg. be back later if you still have questions.
anonymous
  • anonymous
I understand now! Thank you for bearing with me (haha) and helping me through all the steps!!!
mathmate
  • mathmate
You're welcome! :)

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