Question

Solve for xThere are two potential roots (Picture has the whole question )

Answer 1

Given the logarithmic equation:

`\log _5x+\log _5(x-2)=2`

Let's solve for the two potential roots, A and B, where A ≤ B.

Apply the rules of loagarithm:

`\log a+\log b=\log ab`

Thus, we have:

`\log _5x(x-2)=2`

Solving further:

`\begin{gathered} 5^(\log _5x(x-2))=5^2 \\ \\ x(x-2)=25 \end{gathered}`

Subtract 25 from both sides of the equation:

`\begin{gathered} x(x-2)-25=25-25 \\ \\ x(x-2)-25=0 \end{gathered}`

Expand using distributive property:

`\begin{gathered} x(x)+x(-2)-25=0 \\ \\ x^2-2x-25=0 \end{gathered}`

Now solve using the quadratic formula:

`\frac{-b\pm\sqrt[]{b^2-4ac}}{2b}`

To find the values of a, b, and c, apply the general quadratic equation:

`\begin{gathered} ax^2+bx+c \\ \\ x^2-2x-25 \end{gathered}`

Where:

a = 1

b = -2

c = -25

Thus, we have:

`\begin{gathered} \frac{-(-2)\pm\sqrt[]{-2^2-4(1)(-25)}}{2(1)} \\ \\ x=\frac{2\pm\sqrt[]{4+100}}{2} \\ \\ x=\frac{2\pm\sqrt[]{104}}{2} \\ \\ x=\frac{2\pm\sqrt[]{26\ast4}}{2} \\ \\ x=\frac{2\pm2\sqrt[]{26}}{2} \\ \\ x=(2)/(2)\pm\frac{2\sqrt[]{26}}{2} \\ \\ x=1\pm\sqrt[]{26} \\ \\ x=1-\sqrt[]{26},\text{ 1+}\sqrt[]{26} \\ \\ x=-4.1,6.1 \end{gathered}`

Hence, we have the values of A and B:

A = -4.099

B = 6.099

A and B are actually roots.

ANSWER:

A = -4.1

B = 6.1

Yes, A is actually a root

Yes, B is actually a root.