Question

Solve Logarithm Equation .
(a) log(x-2)-logx=-1

Answer 1

I would first write the left as one log:
log[(x-2)/x] = -1 (using the properties of logs)
Then write it as an exponent:
10^-1 = (x-2)/x

Answer 2

\[\log(x-2)/x=-1

\[(x-2)/x=e^-1\]
\[(x-2)*e=x\]

\[e*x-e*2=x\]
e*x-x-x*2=0

Answer 3

.1 = (x-2)/x
.1x = x-2
-.9x=-2
x = 20/9

Answer 4

its -1 t bates not +1 so how u write this 1 = (x-2)/x................ ???

Answer 5

and\[\log _{e ^{(x-1)/x}}
its the from of log

Answer 6

THE ANSWER IS x=20/9

Answer 7

logx/log2-logx=-1
logx-log2logx/log2=-1
logx-log2logx=-log2
logx(1-log2)=-log2
logx=-log2/(1-log2)
x=2.22
=20/9

Answer 8

10^(-1) = 0.1
which I used in my solution

Answer 9

how is it ???its can't be cause \[\log _{e} \] or\[\log _{10}\] log base here u use 10 so u can write this problem as log_e and prosed

Answer 10

anytime you see logx it's always assumed to be base 10. You could change the base to e but that would add extra steps

Answer 11

but how can i learn that where i will use lag_e or log_10 sorry for my mistake sir T Bates i'm really sorry

Answer 12

That's alright. Anytime you see log(x) it's base 10, ln(x) is base e.

Answer 13

ok but when we use 2.303*ln with that time we shall write log why can u explain me sir

Answer 14

I'm not sure what the question is... 2.303*ln (???) anytime you have a logarithm you need something inside the logarithm. But in this case this would be base e as ln is always base e