Question

Solve for x.
Log(x+2)+log(x-4)=log55

Answer 1

ok
log 2 + log x = log 3
or, log x = log 3 - log 2
or, log x = log(3/2)
so, x = 3/2

Eq 2 :

log 2 + log x = 3
or, log(2 * x) = 3
or, 2*x = 10 ^ 3
or, x = (10 ^ 3 )/ 2
so , x = 500

Answer 2

ok now do this one help u

Answer 3

No are you reading th question right

Answer 4

yes i am

Answer 5

rewrite it then but better

Answer 6

That is a clear as it gets. The numbers your using are not right

Answer 7

log(x+2)+ log(x-4) = log 55
Using the log rule..
\[\large \log_b(mn) = \log_b M + \log_b N\]
\[\large \log (x+2)(x-4) = \log 55\]

Answer 8

I know all the rules but that is not helping in this case.

Answer 9

@Bearcaesar the reason @TheRealMeeeee 's solutions are wrong is that because he is only copy and pasting it from google. lol

Answer 10

I'm getting to that..
you can eliminate it like this:
\[\large \frac{ \log (x+2)(x-4) }{ \log(55) } = \frac{ \log 55 }{ \log 55 }\]

Answer 11

The resulting equation is..
\[\large \frac{ (x+2)(x-4) }{ 55 } = 1\]
Solve it algebraically from here.

Answer 12

Thanks GirlByte. I was like what is going on.

Answer 13

Thanks so much shamil98

Answer 14

\[\large (x+2)(x-4) = 55\]
\[\large x^2 -2x - 8 = 55\]
and so on,
and no problem :)

Answer 15

Wait but that does not give me the answer because I did it that way

Answer 16

It does..
\[\huge x^2 - 2x - 63 = 0\]

Answer 17

Believe me I entered that answer at least eight times. I am supposed to have x equals something.

Answer 18

Your supposed to solve for x after wards ....
\[\huge (x+7)(x-9) = 0\]
\[\huge x = 9 , x = -7\]
x = -7 back into the original equation this solution does not work
so your only solution is
\[\Huge x=9\]

Answer 19

Ohhhh makes sense

Answer 20

mhm.