Minimum Logarithmic Expression
Find the minimum value of
\[17 \log_{30} x - 3 \log_x 5 + 20 \log_x 15 - 3 \log_x 6 + 20 \log_x 2\]for $ x > 1 $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
We can write
\begin{align*}
&17 \log_{30} x - 3 \log_x 5 + 20 \log_x 15 - 3 \log_x 6 + 20 \log_x 2 \\
&= 17 \log_{30} x - \log_x 5^3 + \log_x 15^{20} - \log_x 6^3 + \log_x 2^{20} \\
&= 17 \log_{30} x + \log_x \frac{15^{20} \cdot 2^{20}}{5^3 \cdot 6^3} \\
&= 17 \log_{30} x + \log_x (2^{17} \cdot 3^{17} \cdot 5^{17}) \\
&= 17 \log_{30} x + 17 \log_x 30 \\
&= 17 \left( \log_{30} x + \frac{1}{\log_{30} x} \right).\end{align*}By AM-GM,
\[\log_{30} x + \frac{1}{\log_{30} x} \ge 2,\]so $ 17 \left( \log_{30} x + \frac{1}{\log_{30} x} \right) \ge 34 $. Equality occurs when $ x = 30, $ so the minimum value is $ \boxed{34} $.
Freeflows