solve for x:
log(5)(11-6x) = log(5)(1-x)

Question

solve for x:
log(5)(11-6x) = log(5)(1-x)

nnesha · Answer

hint: $$\huge\rm log_b x = \log_b y$$
$$\huge\rm\cancel { log_b} x = \cancel{\log_b} y$$
 x=y

anonymous · Answer

Ohhh... X = 2 right?

anonymous · Answer

Or is it no solution?

pawanyadav · Answer

Is 5 it's base

nnesha · Answer

sorry i forgot :(

anonymous · Answer

Yes 5 is the base

anonymous · Answer

That's ok Nnesha. Thanks for trying to help :)

nnesha · Answer

no i'm saying i forgot i ws helping u :P

anonymous · Answer

Oh hahaha

nnesha · Answer

alright so what was your first step ?? :-)

anonymous · Answer

I would cross out the bases because they are the same right?

nnesha · Answer

yep right and then simple algebra!!! :-)

pawanyadav · Answer

16-6x=1-x
15=5x
x=?

nnesha · Answer

and yes you got it right :-)

mathmate · Answer

Hint: dom    log(x) = (0,inf)

anonymous · Answer

Oh this is a tricky no solution problem isn't it....

mathmate · Answer

... because?

anonymous · Answer

The base is 5 instead of 10?

mathmate · Answer

not exactly.  Use the hint!

anonymous · Answer

Oh! I would have no 'y' value right?

nnesha · Answer

nope when there is just log then u have to suppose 10 base

mathmate · Answer

actually, the hint applies to any base, so it was not indicated.

anonymous · Answer

Oh ok, so the answer would just be x = 2?

mathmate · Answer

want further hint?
@lehmad

anonymous · Answer

yes please :p

nnesha · Answer

i'm confused 
do you have to solve for x right ?

mathmate · Answer

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 · Answer

@Nnesha  yes, the solution x=2 is correct for the algebraic part, but not correct for the problem given!

anonymous · Answer

I have to solve for the problem given @Nnesha

anonymous · Answer

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 · Answer

$log_5(11-6x)=log_5(-1)$
so x=2 is not an admissible solution because log(-1) is outside the domain of log.

mathmate · Answer

* -1 is outside the domain of log.

mathmate · Answer

sorry, gtg. be back later if you still have questions.

anonymous · Answer

I understand now! Thank you for bearing with me (haha) and helping me through all the steps!!!

mathmate · Answer

You're welcome!  :)