Help with solving log

[justan4cat]

I'm in my last week of this online class. You guys have been sp helpful! I'm sorry if some of my questions were dumb questions, but I had nobody else to ask. Thanks for not poking fun. I have 4 problems left and I'm done with this course. This one, I'm a bit stuck on what to do first.
I know that log[sub:1a49q0jm]a[/sub:1a49q0jm]M + log[sub:1a49q0jm]a[/sub:1a49q0jm]N = log[sub:1a49q0jm]a[/sub:1a49q0jm] (MN) so...

Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this.

log[sub:1a49q0jm]8[/sub:1a49q0jm] (x + 4) + log[sub:1a49q0jm]8[/sub:1a49q0jm] (x + 2) = log[sub:1a49q0jm]8[/sub:1a49q0jm]48

should my first step be to factor the left?

Log[sub:1a49q0jm]8[/sub:1a49q0jm] (x+4)(x+2) = log48/log8

 

[Mrspi]

I'm in my last week of this online class. You guys have been sp helpful! I'm sorry if some of my questions were dumb questions, but I had nobody else to ask. Thanks for not poking fun. I have 4 problems left and I'm done with this course. This one, I'm a bit stuck on what to do first.
I know that log[sub:2ryyc04p]a[/sub:2ryyc04p]M + log[sub:2ryyc04p]a[/sub:2ryyc04p]N = log[sub:2ryyc04p]a[/sub:2ryyc04p] (MN) so...

Solve. Where appropriate, include approximations to the nearest thousandth. If no solution exists, state this.

log[sub:2ryyc04p]8[/sub:2ryyc04p] (x + 4) + log[sub:2ryyc04p]8[/sub:2ryyc04p] (x + 2) = log[sub:2ryyc04p]8[/sub:2ryyc04p]48

should my first step be to factor the left?

Log[sub:2ryyc04p]8[/sub:2ryyc04p] (x+4)(x+2) = log48/log8Almost!!

log[sub:2ryyc04p]8[/sub:2ryyc04p] (x + 4)(x + 2) = log[sub:2ryyc04p]8[/sub:2ryyc04p] 48

Now...

if log[sub:2ryyc04p]b[/sub:2ryyc04p] a = log[sub:2ryyc04p]b[/sub:2ryyc04p] c, then a = c

Since

log[sub:2ryyc04p]8[/sub:2ryyc04p] (x + 4)(x + 2) = log[sub:2ryyc04p]8[/sub:2ryyc04p] 48

It must be true that

(x + 4)(x + 2) = 48

Solve that quadratic equation.....

 

[justan4cat]

I think this is right.


(x+4)(x+2) = 48
X[sup:3qsg5sil]2[/sup:3qsg5sil]+2x+4x+8=48
X[sup:3qsg5sil]2[/sup:3qsg5sil]+6x=40
X[sup:3qsg5sil]2[/sup:3qsg5sil]+6x+9=49
(x+3)[sup:3qsg5sil]2[/sup:3qsg5sil]=49
X+3=?49
X+3=7
X=4

 

[Deleted member 4993]

I think this is right.


(x+4)(x+2) = 48
X[sup:t1hjxlgo]2[/sup:t1hjxlgo]+2x+4x+8=48
X[sup:t1hjxlgo]2[/sup:t1hjxlgo]+6x=40
X[sup:t1hjxlgo]2[/sup:t1hjxlgo]+6x+9=49
(x+3)[sup:t1hjxlgo]2[/sup:t1hjxlgo]=49
±(x+3) =?49 <<< You should get two values for 'x' - then you need check if both of those are valid.
X+3=7
X=4

 

[justan4cat]

Check:
x=4
Log[sub476a35z]8[/sub476a35z]8 + log[sub476a35z]8[/sub476a35z]6 = log[sub476a35z]8[/sub476a35z]48
1+log6/log8 = log48/log8
1.8617=1.8617
Or
x=-10
Log[sub476a35z]8[/sub476a35z]-6 + log[sub476a35z]8[/sub476a35z]-8 = log[sub476a35z]8[/sub476a35z]48
Log-6-1=log48/log8
Can’t do log of -6
Answer is X=4.

 

[Deleted member 4993]

Check:
x=4
Log[sub:fays83fy]8[/sub:fays83fy]8 + log[sub:fays83fy]8[/sub:fays83fy]6 = log[sub:fays83fy]8[/sub:fays83fy]48
1+log6/log8 = log48/log8
1.8617=1.8617
Or
x=-10
Log[sub:fays83fy]8[/sub:fays83fy]-6 + log[sub:fays83fy]8[/sub:fays83fy]-8 = log[sub:fays83fy]8[/sub:fays83fy]48
Log-6-1=log48/log8
Can’t do log of -6
Answer is X=4. <<< Correct