not sure if i got this right. but i got 1

$$\log_{6} (x+3)+\log_{6} (x-2)=1$$

Question

not sure if i got this right.  but i got 1

$$\log_{6} (x+3)+\log_{6} (x-2)=1$$

mathslover · Answer

$$\log_6(x+3)+log_6(x-2)=log_6(x+3)(x-2)$$ were there any values given for x ?

anonymous · Answer

the only equation i was given is this:$$\log_{6} (x+3)+\log_{6} (x-2)=1$$

mathslover · Answer

ok so u have to prove this ?

anonymous · Answer

i just have to solve it.

anonymous · Answer

use log properties

anonymous · Answer

to combine into simpler equation

anonymous · Answer

the answer i ended up with was 1 but I'm not sure if i got it right

mathslover · Answer

yes ! :
$$\log_a(c)+log_a(b)=log_a(bc)$$

anonymous · Answer

did you get x^2+x-7=0 as your quadratic?

mathslover · Answer

how will u get 0 right up there ?

mathslover · Answer

anonymous · Answer

lol its log base 6

anonymous · Answer

nvm

anonymous · Answer

so what i did was combine them and got log (base 6) x^2+x-6)=1

anonymous · Answer

x^2+x-12=0?

anonymous · Answer

and solve x

anonymous · Answer

and then plug x back into original equation

anonymous · Answer

nvm! i mixed the signs up and just realized what i did! i didn't get 12 and i got 0

anonymous · Answer

excluding values that makes log negative

anonymous · Answer

thank you :)

anonymous · Answer

so if you got 1 then ur answer is wrong

anonymous · Answer

yeah i know. i realized what i did wrong.

anonymous · Answer

log(1-2)=log(-1) which is not true

anonymous · Answer

i got 4 for my answer