`(1000)^x  = 3`

`rArr  log[(1000)^x] = log 3`

`rArr  x log 1000 =  log 3`

`rArr  x log(10^3) = log 3`

`rArr  3 x log 10 =  log 3`

`rArr   3x = log 3        rArr   x = 0.477/3     = 0.159.`

`log_10 80` = `log_10 (8 xx 10)`

=`log_10 8 + log_10 10`

=` log_10 (2^3) + 1`

=`3 log _10 2 + 1`

= (3 x 0.3010 + 1) = 1.9030.

`log_10 5  = log_10  (10/2)`

=` log_10 10 - log_10 2`

=`1 - log_10 2`

= (1 - 0.3010 ) =0. 6990

`log_2 10` =  `(1)/(log_10 2)`

= `1/03010` = `10000/3010`

=` 1000/301`

Log 27 = 1.431.

`rArr   log(3^3) = 1.431 `        `rArr  3 log 3 = 1.431`

`rArr   log 3 = 0.477`

`:.`      log 9 = `log(3^2)` = 2 log3 = (2 x 0.477) = 0.954

`a = b^x , b = c^y , c = a^z `

`rArr  x = log_b a, y = log_c b , z = log_a c`

`rArr    xyz = (log_b a) xx (log_c b) xx (log_a c)`

`rArr     xyz = ((log a)/(log b) xx (log b)/(log c) xx (log c)/(log a))` = 1

`log_10 (1/70)`

=`log_10 1 - log_10 70`

=`log_10 (7 xx 10)`

=` - (log_10 7 + log _10 10`

= - (a + 1)

Given expression = `log_x (p/q) + log_x (q/r) + log_x (r/p)`

=`log_x (p/q xx q/r xx r/p)`

=`log_x 1 = 0`.

Given expression = log`((a^2)/(bc) xx (b^2)/(ac) xx (c^2)/(ab))`

= log1 = 0

Given expression 

= `(1)/(log_a bc + log_a a) + (1)/(log_b ca + log_b b) + (1)/(log_c ab + log_c c)`

=`(1)/(log_a (abc)` + `(1)/(log_b (abc)` +` (1)/(log_c (abc)`

=`log_(abc) a  + log _(abc) b + log_(abc)  c`

=`log_(abc) abc ` = 1