Solve for x log base 5 of x+ log base 5 of 2x+3=1

$\log_{5} (x) + \log_{5} (2 x + 3) = 1$

Step 1

Simplify the left side.

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

$\log_{5} (x (2 x + 3)) = 1$

Simplify by multiplying through.

Apply the distributive property.

$\log_{5} (x (2 x) + x \cdot 3) = 1$

Reorder.

Rewrite using the commutative property of multiplication.

$\log_{5} (2 x \cdot x + x \cdot 3) = 1$

Move $3$ to the left of $x$ .

$\log_{5} (2 x \cdot x + 3 \cdot x) = 1$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\log_{5} (2 (x \cdot x) + 3 \cdot x) = 1$

Multiply $x$ by $x$ .

$\log_{5} (2 x^{2} + 3 \cdot x) = 1$

$\log_{5} (2 x^{2} + 3 x) = 1$

Step 2

Rewrite $\log_{5} (2 x^{2} + 3 x) = 1$ 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 $b^{y} = x$ .

$5^{1} = 2 x^{2} + 3 x$

Step 3

Solve for $x$ .

Rewrite the equation as $2 x^{2} + 3 x = 5^{1}$ .

$2 x^{2} + 3 x = 5$

Subtract $5$ from both sides of the equation.

$2 x^{2} + 3 x - 5 = 0$

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 $a \cdot c = 2 \cdot - 5 = - 10$ and whose sum is $b = 3$ .

Factor $3$ out of $3 x$ .

$2 x^{2} + 3 (x) - 5 = 0$

Rewrite $3$ as $- 2$ plus $5$

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$2 x (x - 1) + 5 (x - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 1$ .

$(x - 1) (2 x + 5) = 0$

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

$x - 1 = 0$

$2 x + 5 = 0$

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.

$x = 1$

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

The final solution is all the values that make $(x - 1) (2 x + 5) = 0$ true.

$x = 1, - \frac{5}{2}$

Step 4

Exclude the solutions that do not make $\log_{5} (x) + \log_{5} (2 x + 3) = 1$ true.

$x = 1$

Solve for x log base 5 of x+ log base 5 of 2x+3=1

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