Solve for x log base x+5 of 2x^2+1=2

$\log_{x + 5} (2 x^{2} + 1) = 2$

Step 1

Rewrite $\log_{x + 5} (2 x^{2} + 1) = 2$ 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 $b^{y} = x$ .

${(x + 5)}^{2} = 2 x^{2} + 1$

Step 2

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $2 x^{2}$ from both sides of the equation.

${(x + 5)}^{2} - 2 x^{2} = 1$

Simplify each term.

Rewrite ${(x + 5)}^{2}$ as $(x + 5) (x + 5)$ .

$(x + 5) (x + 5) - 2 x^{2} = 1$

Expand $(x + 5) (x + 5)$ using the FOIL Method.

Apply the distributive property.

$x (x + 5) + 5 (x + 5) - 2 x^{2} = 1$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 5 (x + 5) - 2 x^{2} = 1$

Apply the distributive property.

$x \cdot x + x \cdot 5 + 5 x + 5 \cdot 5 - 2 x^{2} = 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 5 + 5 x + 5 \cdot 5 - 2 x^{2} = 1$

Move $5$ to the left of $x$ .

$x^{2} + 5 \cdot x + 5 x + 5 \cdot 5 - 2 x^{2} = 1$

Multiply $5$ by $5$ .

$x^{2} + 5 x + 5 x + 25 - 2 x^{2} = 1$

Add $5 x$ and $5 x$ .

$x^{2} + 10 x + 25 - 2 x^{2} = 1$

Subtract $2 x^{2}$ from $x^{2}$ .

$- x^{2} + 10 x + 25 = 1$

Subtract $1$ from both sides of the equation.

$- x^{2} + 10 x + 25 - 1 = 0$

Subtract $1$ from $25$ .

$- x^{2} + 10 x + 24 = 0$

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 10 x + 24$ .

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 10 x + 24 = 0$

Factor $- 1$ out of $10 x$ .

$- (x^{2}) - (- 10 x) + 24 = 0$

Rewrite $24$ as $- 1 (- 24)$ .

$- (x^{2}) - (- 10 x) - 1 \cdot - 24 = 0$

Factor $- 1$ out of $- (x^{2}) - (- 10 x)$ .

$- (x^{2} - 10 x) - 1 \cdot - 24 = 0$

Factor $- 1$ out of $- (x^{2} - 10 x) - 1 (- 24)$ .

$- (x^{2} - 10 x - 24) = 0$

Factor.

Factor $x^{2} - 10 x - 24$ 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 $- 24$ and whose sum is $- 10$ .

$- 12, 2$

Write the factored form using these integers.

$- ((x - 12) (x + 2)) = 0$

Remove unnecessary parentheses.

$- (x - 12) (x + 2) = 0$

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

$x - 12 = 0$

$x + 2 = 0$

Set $x - 12$ equal to $0$ and solve for $x$ .

Set $x - 12$ equal to $0$ .

$x - 12 = 0$

Add $12$ to both sides of the equation.

$x = 12$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

The final solution is all the values that make $- (x - 12) (x + 2) = 0$ true.

$x = 12, - 2$

Solve for x log base x+5 of 2x^2+1=2

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