how do I solve log (x) - 1 = log (x-1)?

Question

how do I solve (with the steps)

log (x) - 1 = log (x-1)

both logs are in base 8

Answer 1

First remember

1 = log 8

Then we have

log (x) - 1 = log (x-1)
log (x) - log 8 = log (x-1)
log (x/8) = log (x-1)

Now exponentiate.

x/8 = x - 1
x = 8(x - 1) = 8x - 8
8 = 7x
8/7 = x
x = 8/7

Answer 2

Log(x) - 1 = Log(x-1)
Log(x) - Log(x-1) = 1
Log( x/(x-1) ) = 1 = Log(e) (Or Log(10), depending on base)
x/(x-1) = a (a is e or 10, depending)
x = ax - a
a = (a-1)x
a/(a-1) = x = e/(e-1) (if base e), or = 10/9 (if base 10)

Oh, it's 8/7 if base 8. Okay.

Answer 3

you subtract log(x-1) from each side.
log(x)-log(x-1)-1=0
then you combine the two logorithms
log(x/(x-1))-1=0
you add one to each side
log(x/(x-1))=1
and then since it's base 8 you can take 8 to the log(x/(x-1)) power and 8 to the 1st power. then it cancels out the logorithm
x/(x-1)=8
solving out that equation you get
x=8/7

Answer 4

log10^2x+1)-log10^(x-2=log10^1O NOTE LOG10^10=1 all bases cut by 10 2x+1/x-2=10 2x+1=10(x-2 2x+1=10x-20 -8x=-21 x=21/8

Answer 5

Just wrap it back: x/8 = x-1, hence x = 8/7;

Answer 6

logb(m/n) = logb(m) – logb(n) is the rule....

so you have

log (x/(x-1)) = 1
then 8^1 = x/(x-1)

so 8x -8=x
7x=8

x=8/7

...i think