Find other values of a, b, and c so that log(a) + log(b) + log(c) = log(a + b + c).

Step-by-step explanation:

Given :

log(a) + log(b) + log(c) = log(a + b + c)

We know that by property

log(a) + log(b) + log(c) = log(abc)

Therefore,

log(abc) = log(a + b + c)

Thus,

(abc) = (a + b + c)

Thus when a = b = c = 0,

         then (abc) = (a + b + c)

                  0 = 0

and when a = 1, b = 2, c = 3,

       then (abc) = (a + b + c)

               (1 x 2 x 3 ) = (1 + 2 + 3)

                    6 = 6

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

b

2

−

b

b

3

+

7

b

2

+

14

b

+

6

%

−

3

b

3

+

7

b

2

+

14

b

+

6

%

step-by-step explanation: