Question

log_6(x)+log_6(x+1)=1

Answer 1

Combine the logs on the left like this
\[\log_b(N)+\log_b(M)=\log_b(N\times M)\]

Answer 2

Yeah I got log_6(x)(x+1)=1 so far

Answer 3

Now use the definition of a log
\[\log_bz=y\quad\iff\quad z=b^y\]

Answer 4

can you do this part?

Answer 5

That means do the 6 to the power both sides

Answer 6

yes,

Answer 7

@MikeB55 your turn now :)

Answer 8

log_6(x^2+x)=log_6 6 @MikeB55

Answer 9

you'll get a quadratic which has two solutions,
only one of these solutions solves the original equation

Answer 10

Looks like Mike's not here, let's wait for his return :)

Answer 11

Hey sorry everyone

Answer 12

I'm at x^2+x=6

Answer 13

So x=2

Answer 14

Yep

Answer 15

Thanks everyone

Answer 16

no problem

Answer 17

that's right!, did you also get the number that solves the quadratic but not the log equation?
do you know why this solution doesn't work?

Answer 18

I assume he already knows it :)

Answer 19

Yeah I got it!

Answer 20

why -3 is not accepted @Unkle

Answer 21

Because log of negative is undefined

Answer 22

Negative numbers can't be used to solve a log equation

Answer 23

ok

Answer 24

exactly