If the equation $y=5(1.1)^{x}$ is plotted on a semilog scale graph, then the linearized equation will be $\log (y)=m x+b$ where
\[
m=
\]
\[
b=
\]

Step 1 ：The given equation is in the form of an exponential function, $y = ab^x$. When this is plotted on a semilog scale, it becomes a straight line. The slope of this line is the logarithm of the base of the exponential function, and the y-intercept is the logarithm of the coefficient of the exponential function.

Step 2 ：In this case, the base of the exponential function is 1.1 and the coefficient is 5. Therefore, the slope of the line, m, is $\log(1.1)$ and the y-intercept, b, is $\log(5)$.

Step 3 ：Calculating these values, we find that $m = \log(1.1) \approx 0.0414$ and $b = \log(5) \approx 0.699$.

Step 4 ：Therefore, the linearized equation of $y=5(1.1)^{x}$ on a semilog scale graph is $\log (y)=0.0414x+0.699$. So, $m = \boxed{0.0414}$ and $b = \boxed{0.699}$.