Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 457: 30

$1.8$

Apply logarithmic property : $\log m+ \log n=\log (mn)$ We can write the equation as: $ pH=-\log [1.6 \times 10^{-2}] =-[\log 1.6 + \log 10^{-2}]$ Now, apply logarithmic property : $\log a^ b=b \log a$ Therefore, our answer is: $-[\log 1.6 - 2 \log 10]=-0.204-2= 1.8$