There is a big difference between
\log{A}
(line 3 of your code) and
\logA
(line 4 of your code), but they're both wrong and it should be \log A
in both cases.
Actually, \log{A}
and \log A
produce the same output, but it wouldn't be the same for \log{(A+B)
and \log (A+B)
(check it).
You should use amsmath
and input your equation as
\begin{equation}
S =
\begin{cases}
1280Q^{0.46}(1.43-0.26\log A) & \text{if $Q<2\,\mathrm{in}$} \\
1958e^{-0.055}Q(1.43-0.26\log A) & \text{if $Q\geq 2\,\mathrm{in}$}
\end{cases}
\end{equation}
(in mathematics, multiplication is very rarely denoted by an asterisk).
Full code:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
S =
\begin{cases}
1280Q^{0.46}(1.43-0.26\log A) & \text{if $Q<2\,\mathrm{in}$} \\
1958e^{-0.055}Q(1.43-0.26\log A) & \text{if $Q\geq 2\,\mathrm{in}$}
\end{cases}
\end{equation}
\end{document}

\logA
. Replace it by\log A
, i.e. add a space. But you really may want to have a look at some intro, and use e.g.\begin{cases} ... \end{cases}
and\text
instead of\mbox
and so on.\log{A}
and\logA
: both should be\log A
(no braces and a space).