Find x, if :

2 + log x = log 45 - log 2 + log 16 - 2 log 3.

Given: 2 + log x = log 45 - log 2 + log 16 - 2 log 3

⇒ log $10^2$ + log x = log (9 x 5) - log 2 + log $(2^4)$ - 2 log 3

⇒ log $10^2$ + log x = log 9 + log 5 - log 2 + 4 log 2 - 2 log 3

⇒ log $10^2$ + log x = log $(3^2)$ + log 5 + 3 log 2 - 2 log 3

⇒ log $10^2$ + log x = 2 log 3 + log 5 + 3 log 2 - 2 log 3

⇒ log $10^2$ + log x = log 5 + 3 log 2

⇒ log x = log 5 + 3 log 2 - log $10^2$

⇒ log x = log 5 + 3 log 2 - log $(5 \times 2)^2$

⇒ log x = log 5 + 3 log 2 - (log $5^2$ + log $2^2$)

⇒ log x = log 5 + 3 log 2 - 2log 5 - 2log 2

⇒ log x = log 2 - log5

⇒ log x = log $\dfrac{2}{5}$

⇒ x = $\dfrac{2}{5}$

⇒ x = 0.40

Hence, the value of x = 0.40.