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

- anonymous

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

- chestercat

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- Nnesha

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

- anonymous

Ohhh... X = 2 right?

- anonymous

Or is it no solution?

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## More answers

- Pawanyadav

Is 5 it's base

- Nnesha

sorry i forgot :(

- anonymous

Yes 5 is the base

- anonymous

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

- Nnesha

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

- anonymous

Oh hahaha

- Nnesha

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

- anonymous

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

- Nnesha

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

- Pawanyadav

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

- Nnesha

and yes you got it right :-)

- mathmate

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

- anonymous

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

- mathmate

... because?

- anonymous

The base is 5 instead of 10?

- mathmate

not exactly. Use the hint!

- anonymous

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

- Nnesha

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

- mathmate

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

- anonymous

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

- mathmate

want further hint?
@lehmad

- anonymous

yes please :p

- Nnesha

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

- 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

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

- anonymous

I have to solve for the problem given @Nnesha

- 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

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

- mathmate

* -1 is outside the domain of log.

- mathmate

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

- anonymous

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

- mathmate

You're welcome! :)

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