\section*{Problem 6 - 041016}

a) Calculate an integer approximation for

\[
x=\frac{\lg 5}{\lg 3-\lg 2}
\]

without using any computational aids.

b) Is this approximate value smaller or larger than the exact value?
}

Used formulas (known from school lessons):

\[
\begin{array}{r}
\log _{b} \frac{u}{v}=\log _{b} u-\log _{b} v \\
\log _{c} a=\frac{\log _{b} a}{\log _{b} c} \\
\log _{b} a=x \Leftrightarrow b^{x}=a
\end{array}
\]

According to (1), we have

\[
\lg 3-\lg 2=\lg \frac{3}{2}, \quad \text { i.e. } \quad x=\frac{\lg 5}{\lg \frac{3}{2}}
\]

According to (2) and (3), we have:

\[
x=\log _{\frac{3}{2}} 5 \Leftrightarrow\left(\frac{3}{2}\right)^{x}=5
\]

By trying out integer values for \(x\), we get:

\[
\left(\frac{3}{2}\right)^{1}=1.5 ; \quad\left(\frac{3}{2}\right)^{2}=2.25 ; \quad\left(\frac{3}{2}\right)^{3}=3.375 ; \quad\left(\frac{3}{2}\right)^{4}=5.0625 ; \quad\left(\frac{3}{2}\right)^{5}=7.59375
\]

Therefore, 4 is the best integer approximation for \(x\), as it is closest to the solution of the equation \((=5)\). Since \(\left(\frac{3}{2}\right)^{4}>5\), the approximate value \(x=4\) is greater than the actual solution.