solve log (x-1)-log8 x=1

Hi Omi

I agree, but you forgot the log₈(x) term.  

ln(x-1)/ln(8) - ln(x)/ln(8) = 1 

See Guest's development in #4 above.

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

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

Hallo guest

Hmmm

\(\log_8{(x-1)}-\log_8{x}=1 \rightarrow\log_8{(\frac{x-1}{x})}=1\rightarrow\frac{x-1}{x}=8^1\rightarrow\frac{x-1}{x}=8\\\\ \rightarrow x-1=8x\rightarrow7x=-1\rightarrow x=-\frac{1}{7}\)

Slightly odd, as the original individual logarithms would only have complex values, though the combined log would be real (the imaginary parts cancel out).

Solve for x:
(log(x-1))/(log(8))-(log(x))/(log(8)) = 1

Rewrite the left hand side by combining fractions. (log(x-1))/(log(8))-(log(x))/(log(8))  =  (log(x-1)-log(x))/(log(8)):
(log(x-1)-log(x))/(log(8)) = 1

Multiply both sides by log(8):
log(x-1)-log(x) = log(8)

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

Cancel logarithms by taking exp of both sides:
(x-1)/x = 8

Multiply both sides by x:
x-1 = 8 x

Subtract 8 x-1 from both sides:
-7 x = 1

Divide both sides by -7:
Answer: | 
| x = -1/7(assuming a complex-valued Logarithm)

Hallo Alan, I think in such a way

???