Given:
[tex]\log _7(x^2+6)-\log _7(x+2)=1[/tex]To find: Solve all values of x?
Explanation:
We know the Logarithm quotient rule
[tex]\log _b(\frac{x}{y})=\log _b(x)-\log _b(y)[/tex]We can write as
[tex]\begin{gathered} \log _7(x^2+6)-\log _7(x+2)=1 \\ \\ By\text{ using the rule } \\ \\ \log _7(\frac{x^2+6}{x+2})=1 \\ \\ (\frac{x^2+6}{x+2})=7^1 \\ \\ \frac{x^2+6}{x+2}=7 \end{gathered}[/tex]Here cross multiply
[tex]\begin{gathered} x^2+6=7\times(x+2) \\ \\ x^2+6=7x+14 \\ \\ x^2+6-7x-14=0 \\ \\ x^2-7x-8=0 \end{gathered}[/tex]Split middle term
[tex]\begin{gathered} x^2-8x+x-8=0 \\ \\ x(x-8)+1(x-8)=0 \\ \\ \text{take (x-8) co}mmeon \\ \\ (x-8).(x+1)=0 \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} x-8=0 \\ x=8 \\ \\ or \\ \\ x+1=0 \\ x=-1 \end{gathered}[/tex]Hence, the values of x = 8 or x = -1.
Thus, the values of x = 8 or x = -1.