The problem asks to find the function that results from transforming $f(x) = \log_{10}x$ by a vertical stretch by a factor of 4, horizontal stretch by a factor of 3, reflection in the y-axis, horizontal translation 5 units to the right, and vertical translation 2 units up.

1. Problem Description

The problem asks to find the function that results from transforming $f(x) = \log_{10}x$ by a vertical stretch by a factor of 4, horizontal stretch by a factor of 3, reflection in the y-axis, horizontal translation 5 units to the right, and vertical translation 2 units up.

2. Solution Steps

The original function is $f(x) = \log_{10}x$.

Vertical stretch by a factor of 4: $4\log_{10}x$.

Horizontal stretch by a factor of 3: $4\log_{10}(x/3)$. Note that a horizontal stretch by a factor of 3 means replacing $x$ by $x/3$. This gives $f(x)=4\log_{10}(x/3)$.

Reflection in the y-axis: $4\log_{10}(-x/3)$.

Horizontal translation 5 units to the right: $4\log_{10}(-(x-5)/3)$.

Vertical translation 2 units up: $4\log_{10}(-(x-5)/3) + 2$.

Simplifying the argument inside the logarithm: $4\log_{10}(\frac{-(x-5)}{3}) + 2 = 4\log_{10}(\frac{-x+5}{3}) + 2$.

An alternative approach is to first take $f(x)=\log_{10}(x)$, then $f(x/(\frac{1}{3}))=\log_{10}(\frac{x}{\frac{1}{3}})=\log_{10}(3x)$.

Reflection in the y-axis means we replace x by -x to get $\log_{10}(-3x)$.

A horizontal translation 5 units to the right means we replace x by (x-5), i.e., $f(x-5)$.

So we have $f(x)=\log_{10}(-3(x-5))$.

Vertical stretch by a factor of 4 gives $4\log_{10}(-3(x-5))$.

Vertical translation 2 units up gives $4\log_{10}(-3(x-5))+2$.

3. Final Answer

$f(x) = 4\log_{10}[-3(x-5)] + 2$