1. ## Log Equation

Log(base6) - log(base 6)(x-1)=1

MY SET UP:

log_6[1/(x-1)] = 1

6^1 = 1/(x-1)

6 = 1/(x-1)

6(x-1) = 1

6x - 6 = 1

6x = 7

x = 7/6

x = 1.2

Right or wrong? If wrong, why?

2. Looks good to me

3. That's correct.

Originally Posted by magentarita
Log(base6) - log(base 6)(x-1)=1

MY SET UP:

log_6[1/(x-1)] = 1

6^1 = 1/(x-1)

6 = 1/(x-1)

6(x-1) = 1

6x - 6 = 1

6x = 7

x = 7/6

x = 1.2

Right or wrong? If wrong, why?

4. Originally Posted by magentarita
Log(base6) - log(base 6)(x-1)=1

MY SET UP:

log_6[1/(x-1)] = 1

6^1 = 1/(x-1)

6 = 1/(x-1)

6(x-1) = 1

6x - 6 = 1

6x = 7

x = 7/6

x = 1.2

Right or wrong? If wrong, why?
It looks like you left something out of the question...

$\log_6\left(\dots\right)-\log_6\left(x-1\right)=1$

Would you fill in the dots?

PS: If it is indeed $\log_6\left(1\right)-\log_6\left(x-1\right)=1$, then your work is correct.

5. Originally Posted by magentarita
Log(base6) - log(base 6)(x-1)=1

MY SET UP:

log_6[1/(x-1)] = 1

6^1 = 1/(x-1)

6 = 1/(x-1)

6(x-1) = 1

6x - 6 = 1

6x = 7

x = 7/6

x = 1.2

Right or wrong? If wrong, why?
1. In the original equation the argument of the first log-term is missing. According to your calculations it seems to be 1.

2. All calculations are correct - but $\dfrac76 \neq 1.2$

3. You can control your work yourself by plugging in your solution into the original equation. If you get a true statement then your work is correct.

....

121b) May I ask you a favour? Sometimes it is very difficult for me to read your questions. Why don't you use simply black colour for writing?

6. ## ok...

I thank both replies. I will use simply black ink from now when asking questions.