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

I need help please and thank you Solve all values of x-example-1
User Amolgautam
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1 Answer

4 votes

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.

User Hymns For Disco
by
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