Let `a >1` be a real number. Then the number of roots equation `a^(2(log)_2x)=15+4x^((log)_2a)` is 2 (b) infinite (c) 0 (d) 1

Explanation

`a^(2log_2^x)=15+4x^(log_2^a)`

`a>1`

`(a^(log_2^x))^2=15+4x^(log_2^a)`

`(x^(log_2^a))^2=15+4x^(log_2^a)`

`x>0`

Let `x^(log_2^a)=t`

`log_2^a` is positive

`t^2=15+4t`

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