The equation 5^1+log_5 cos x= 2.5 has

We have,

$5^{1+\log_5 \cos x}= \frac{5}{2}$ taking log to the base '5' on both sides

$⇒1 + \log_5 \cos x = \log_55-\log_52$

$⇒\log_5 \cos x =-\log_5 2$

so $\cos x=2^{-1}⇒\cos x=\frac{1}{2}$

$⇒x=2n\,π±\frac{π}{3},n∈Z$.