Answer:
Third option
[tex]\log _{10}\left(20x^5\right)-1\\[/tex]
Step-by-step explanation:
[tex]5\log _{10}\left(x\right)+\log _{10}\left(20\right)-\log _{10}\left(10\right)\\\\\mathrm{Apply\:log\:rule}:\\\quad \:a\log _c\left(b\right)=\log _c\left(b^a\right)\\\\5\log _{10}\left(x\right)=\log _{10}\left(x^5\right)\\\\=\log _{10}\left(x^5\right)+\log _{10}\left(20\right)-\log _{10}\left(10\right)\\\\\mathrm{Apply\:log\:rule}:\\\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)\\\\\log _{10}\left(x^5\right)+\log _{10}\left(20\right)=\log _{10}\left(20x^5\right)\\\\[/tex]
[tex]=\log _{10}\left(x^5\cdot \:20\right)-\log _{10}\left(10\right)\\\\\mathrm{Apply\:log\:rule}:\quad \log _a\left(a\right)=1\\\\=\log _{10}\left(20x^5\right)-1[/tex]
