Answer:
[tex]\large\boxed{x=14}[/tex]
Step-by-step explanation:
[tex]3\log_5(x-10)-\dfrac{1}{2}\log_54=5\log_52\\\\\text{Domain:}\ x-10>0\to x>10\\\\\text{use}\\\\\log_ab^n=n\log_ab\\\\\log_ab-\log_ac=\log_a\left(\dfrac{b}{c}\right)\\========================\\\\\log_5(x-10)^3-\log_54^\frac{1}{2}=\log_52^5\qquad\text{use}\ a^\frac{1}{2}=\sqrt{a}\\\\\log_5(x-10)^3-\log_5\sqrt4=\log_532\\\\\log_5(x-10)^3-\log_52=\log_532\\\\\log_5\dfrac{(x-10)^3}{2}=\log_532\iff\dfrac{(x-10)^3}{2}=32\qquad\text{multiply both sides by 2}\\\\(x-10)^3=64\to x-10=\sqrt[3]{64}\\\\x-10=4\qquad\text{add 10 to both sides}\\\\x=14\in D[/tex]
