Question

I am indesperate need of Q9

Answer 1

2^x = 3^y = 6^(-z) = k (Say)

Well,

2^x = k => x = log_2 (k) = log (k) / log (2)

=> 1/x = log (2) / log (k)

3^y = k => y = log_3 (k) = log (k) / log (3)

=> 1/y = log (3) / log (k)

6^(-z) = k => z = - log_6 (k) = - log (k) / log (6)

=> 1/z = - log (6) / log (k)

Now,

(1/x) + (1/y) + (1/z)

= (log (2) / log (k)) + (log (3) / log (k)) - (log (6) / log (k)) +

= [log (2) + log (3) - log (6)] / log (k)

= log (2 × 3 / 6) / log (k)

= log (1) / log (k)

= 0

Answer 2

Answer:

Sorry

Step-by-step explanation:

1. I don't know
2. sbhdhddbbbdbbddrj