1

$\log_5 (10)\cdot\log_{10} (15)\cdot \log_{15} (20)\cdot\log_{20} (25)$

I thought I could apply $\log_a(a)$ but I am not able to understand how.

Arnaldo
  • 21,342

2 Answers2

2

Hint

$$\log_{a}b=\frac{\log_{c}b}{\log_{c}a}$$

You can, for example, write

$$\log_{5}10=\frac{\log10}{\log5}$$

Arnaldo
  • 21,342
-1

You can apply change of basis (for basis $10$), i.e., $log_ba=\frac{loga}{logb}$. Thus, $$\log_5 (10)\cdot\log_{10} (15)\cdot \log_{15} (20)\cdot\log_{20} (25)=\frac{log10}{log5}\frac{log15}{log10}\frac{log20}{log15}\frac{log25}{log20}=\frac{log25}{log5}=log_525=2$$

Cgomes
  • 1,184