Solving for X: log_8(x + 1) = log_8(2x - 2)

[hollerback1]

Hi, This is how far I got...

Solve for X:

log₈(x+1)=log₈(2x-2)

I guessed x=3, how do you find x?

 

[tkhunny]

Re: Solving for X?!!! logarithm...very challenging

Hi, This is how far I got...

Solve for X:

log₈(x+1)=log₈(2x-2)

I guessed x=3 I dunno how to solve this problem

In my view, this is first, and foremost, a DOMAIN problem. Since the problem statement is in logarithms, there are some necessary restrictions.

x+1 > 0 ==> x > -1
2x-2 > 0 ==> x-1 > 0 ==> x > 1

So, let's remember that x > 1.

log₈(x+1)=log₈(2x-2)
Solve: x+1 = 2x-2

1) Why can I just do that?
2) What is the result?
3) Is that the answer?

 

[hollerback1]

oooo, so do you guess what the answer is after u find the zero?

like, if x>1 then it shud b 3 right?

because, (3)+1=4 and (3)(2)-2= 4

plz tell me :?

 

[tkhunny]

There is no guessing involved.

Solve the equation and get x = 3.
Substitute this value into the ORIGINAL equation and see if it works.

Show ALL your work.

 

[hollerback1]

I did substitute and I got x=3, but I guess/checked it.

How are you sure to find the right number for the substitution?

 

[tkhunny]

You keep missing this part: Solve: x+1 = 2x-2

x+1 = 2x-2

Subtract 'x' from each side

x-x+1 = 2x-x-2
0+1 = x-2
1 = x-2

Add '2' to each side

1+2 = x-2+2
3 = x+0
3 = x <== That's where you get x = 3. No guessing.

 

[hollerback1]

oh, ok thanks alot. I get it now