Solve for x log base 6 of x+5+ log base 6 of x=2

$\log_{6} (x + 5) + \log_{6} (x) = 2$

Step 1

Simplify the left side.

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{6} ((x + 5) x) = 2$

Apply the distributive property.

$\log_{6} (x \cdot x + 5 x) = 2$

Multiply $x$ by $x$ .

$\log_{6} (x^{2} + 5 x) = 2$

Step 2

Rewrite $\log_{6} (x^{2} + 5 x) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$6^{2} = x^{2} + 5 x$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} + 5 x = 6^{2}$ .

$x^{2} + 5 x = 6^{2}$

Raise $6$ to the power of $2$ .

$x^{2} + 5 x = 36$

Subtract $36$ from both sides of the equation.

$x^{2} + 5 x - 36 = 0$

Factor $x^{2} + 5 x - 36$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 36$ and whose sum is $5$ .

$- 4, 9$

Write the factored form using these integers.

$(x - 4) (x + 9) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 9 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 9$ equal to $0$ and solve for $x$ .

Set $x + 9$ equal to $0$ .

$x + 9 = 0$

Subtract $9$ from both sides of the equation.

$x = - 9$

The final solution is all the values that make $(x - 4) (x + 9) = 0$ true.

$x = 4, - 9$

Step 4

Exclude the solutions that do not make $\log_{6} (x + 5) + \log_{6} (x) = 2$ true.

$x = 4$

Solve for x log base 6 of x+5+ log base 6 of x=2

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