Answer :
[tex]\begin{gathered} 6-\log _9(x+10)=5 \\ -\log _9(x+10)=5-6 \\ -\log _9(x+10)=-1 \\ \log _9(x+10)=1 \\ x+10=9^1 \\ x+10=9 \\ x=9-10 \\ x=-1 \end{gathered}[/tex][tex]\begin{gathered} \text{tellme your question please} \\ ok\text{ I will solve it above, give a moment please} \\ I\text{ did it, can you s}e\text{ it?} \end{gathered}[/tex][tex]\begin{gathered} \log _5x=2 \\ applyinbothsideiniversefunctiontolog,whichis5^{\log x}=5^2,\text{ hence} \\ x=5^2 \\ x=25 \end{gathered}[/tex]