Question

hello


logarithms 78 SM
Plz. give correct ans. in detail step by step. ​

Answer 1

Answer:

OPTION B

Step-by-step explanation:

Let us take a variable k such that :

1/4 log₂a = 1/6 log₂b = -1/24 log₂c = log₂k

Then the above expressions can be equated one by one to obtain the value :

1/4 log₂a = log₂k

⇒ log₂a^(1/4) = log₂k

Take antilog both sides :-

⇒ a^(1/4) = k

⇒ a = k⁴ .

1/6 log₂b = log₂k

⇒ log₂b^(1/6) = log₂k

Take antilog both sides :-

⇒ b^(1/6) = k

⇒ b = k⁶

Similarly if we do with log₂c :

- 1/24 log₂c = log₂k

⇒ log₂c^(-1/24) = log₂k

Take antilog :

⇒ c^(-1/24) = k

⇒ c = k^(-24)

⇒ c = 1/k²⁴

a³b²c

⇒ ( k⁴ )³ . ( k⁶ )² / k²⁴

⇒ k¹² . k¹² / k²⁴

⇒ k¹²⁺¹² / k²⁴

⇒ k²⁴ / k²⁴

⇒ 1

The value will be 1 .

Answer 2

The correct answer is option b.