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If (log)k xdot(log)5k=(log)x5,k!=1,k >0,...

If `(log)_k xdot(log)_5k=(log)_x5,k!=1,k >0,` then `x` is equal to `k` (b) 1/5 (c) 5 (d) none of these

A

k

B

`1//5`

C

5

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
B, C

`log_(k) x * log_(5) k=log_(x) 5`
` or(log x log k)/(log k log 5) = log_(x) 5 `
` or (log x )/(log 5) = log _(x) 5`
` or log_(5) x = 1/(log_(5) x)`
` or (log_(5)x)^(2) = 1`
` or log_(5) x = pm 1`
` rArr x = 5^(pm 1) `
` rArr x = 1/5 , 5`
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