Question

I need help please and thank you Solve all values of x

Answer 1

Given:

`\log _7(x^2+6)-\log _7(x+2)=1`

To find: Solve all values of x?

Step-by-step explanation:

We know the Logarithm quotient rule

`\log _b((x)/(y))=\log _b(x)-\log _b(y)`

We can write as

`\begin{gathered} \log _7(x^2+6)-\log _7(x+2)=1 \\ \\ By\text{ using the rule } \\ \\ \log _7((x^2+6)/(x+2))=1 \\ \\ ((x^2+6)/(x+2))=7^1 \\ \\ (x^2+6)/(x+2)=7 \end{gathered}`

Here cross multiply

`\begin{gathered} x^2+6=7*(x+2) \\ \\ x^2+6=7x+14 \\ \\ x^2+6-7x-14=0 \\ \\ x^2-7x-8=0 \end{gathered}`

Split middle term

`\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}`

Therefore,

`\begin{gathered} x-8=0 \\ x=8 \\ \\ or \\ \\ x+1=0 \\ x=-1 \end{gathered}`

Hence, the values of x = 8 or x = -1.

Thus, the values of x = 8 or x = -1.