Question

plzzz explain each and every steps and answer fast

Answer 1

Answer:

0

Step-by-step explanation:

Given: 2^(a) = 3^(b) = 6^(-c).

Let 2^a = 3^b = 6^-c = k{∴ some constant}

(i)

2^a = k

Apply log on both sides, we get

⇒ log(2^a) = log k

∴ log aⁿ = n log a.

⇒ a log 2 = log k

⇒ log 2 = log k/a

(ii)

3^b = k

Apply log on both sides, we get

⇒ log(3^b) = log k

⇒ b log 3 = log k

⇒ log 3 = log k/b

(iii)

6^-c = k

Apply log on both sides, we get

⇒ log(6^-c) = log k

⇒ -c log 6 = log k

⇒ log 6 = log k/-c

Now,

Equation (iii) can be written as,

⇒ log 6 = log k/-c

⇒ log(2 * 3) = log k/-c

∴ log(m * n) = log m + log n

⇒ log 2 + log 3 = log k/-c

⇒ log k/a + log k/b = log k/-c

⇒ 1/a + 1/b = -1/c

⇒ 1/a + 1/b + 1/c = 0

⇒ (bc + ca + ab)/abc = 0

⇒ ab + bc + ca = 0

Therefore, the value of ab + bc + ca = 0.

Hope this helps!

Answer 2

Step-by-step explanation:

2^a = 6^-c => 2^a =2^-c * 3^ -c => 2^a * 3^0 = 2^ -c * 3^ -c [as, 3^0 =1}

=> a= -c and 0 = -c [if, a^m * b^k = a^n * b^t then, m=n ,k=t ]

so a= -c = 0 ....(i)

again, 3^b = 6^ -c => 3^b *2^0 = 3^ -c * 2^ -c {2^0 =1]

=> b= -c and 0= -c [as, a^m *b^k = a^n b^t => m=n , k=t ]

so, b= -c =0 ....(ii)

now, LHS = ab+bc+ca = 0+0+0 =0