Solve for x log base 5 of 4x^2+2- log base 5 of x+5=0

$\log_{5} (4 x^{2} + 2) - \log_{5} (x + 5) = 0$

Step 1

Simplify the left side.

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{5} (\frac{4 x^{2} + 2}{x + 5}) = 0$

Factor $2$ out of $4 x^{2} + 2$ .

Factor $2$ out of $4 x^{2}$ .

$\log_{5} (\frac{2 (2 x^{2}) + 2}{x + 5}) = 0$

Factor $2$ out of $2$ .

$\log_{5} (\frac{2 (2 x^{2}) + 2 (1)}{x + 5}) = 0$

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

$\log_{5} (\frac{2 (2 x^{2} + 1)}{x + 5}) = 0$

Step 2

Rewrite $\log_{5} (\frac{2 (2 x^{2} + 1)}{x + 5}) = 0$ 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 $b^{y} = x$ .

$5^{0} = \frac{2 (2 x^{2} + 1)}{x + 5}$

Step 3

Cross multiply to remove the fraction.

$2 (2 x^{2} + 1) = 5^{0} (x + 5)$

Step 4

Simplify $5^{0} (x + 5)$ .

Anything raised to $0$ is $1$ .

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

Multiply $x + 5$ by $1$ .

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

Step 5

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

Subtract $x$ from both sides of the equation.

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

Simplify each term.

Apply the distributive property.

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

Multiply $2$ by $2$ .

$4 x^{2} + 2 \cdot 1 - x = 5$

Multiply $2$ by $1$ .

$4 x^{2} + 2 - x = 5$

Step 6

Move all terms not containing $x$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

$4 x^{2} - x = 5 - 2$

Subtract $2$ from $5$ .

$4 x^{2} - x = 3$

Step 7

Subtract $3$ from both sides of the equation.

$4 x^{2} - x - 3 = 0$

Step 8

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot - 3 = - 12$ and whose sum is $b = - 1$ .

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

$4 x^{2} - (x) - 3 = 0$

Rewrite $- 1$ as $3$ plus $- 4$

$4 x^{2} + (3 - 4) x - 3 = 0$

Apply the distributive property.

$4 x^{2} + 3 x - 4 x - 3 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 x^{2} + 3 x) - 4 x - 3 = 0$

Factor out the greatest common factor (GCF) from each group.

$x (4 x + 3) - (4 x + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 x + 3$ .

$(4 x + 3) (x - 1) = 0$

Step 9

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

$4 x + 3 = 0$

$x - 1 = 0$

Step 10

Set $4 x + 3$ equal to $0$ and solve for $x$ .

Set $4 x + 3$ equal to $0$ .

$4 x + 3 = 0$

Solve $4 x + 3 = 0$ for $x$ .

Subtract $3$ from both sides of the equation.

$4 x = - 3$

Divide each term in $4 x = - 3$ by $4$ and simplify.

Divide each term in $4 x = - 3$ by $4$ .

$\frac{4 x}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 3}{4}$

Divide $x$ by $1$ .

$x = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

Step 11

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 12

The final solution is all the values that make $(4 x + 3) (x - 1) = 0$ true.

$x = - \frac{3}{4}, 1$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{3}{4}, 1$

Decimal Form:

$x = - 0.75, 1$

Solve for x log base 5 of 4x^2+2- log base 5 of x+5=0

Related Questions

Support LCMGCF.com by disabling your adblocker.