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

$\log (5 x + 1) = \log (2 x + 3) + \log (2)$

Step 1

Simplify the right side.

Simplify $\log (2 x + 3) + \log (2)$ .

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

$\log (5 x + 1) = \log ((2 x + 3) \cdot 2)$

Apply the distributive property.

$\log (5 x + 1) = \log (2 x \cdot 2 + 3 \cdot 2)$

Multiply.

Multiply $2$ by $2$ .

$\log (5 x + 1) = \log (4 x + 3 \cdot 2)$

Multiply $3$ by $2$ .

$\log (5 x + 1) = \log (4 x + 6)$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$5 x + 1 = 4 x + 6$

Step 3

Solve for $x$ .

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

Subtract $4 x$ from both sides of the equation.

$5 x + 1 - 4 x = 6$

Subtract $4 x$ from $5 x$ .

$x + 1 = 6$

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

Subtract $1$ from both sides of the equation.

$x = 6 - 1$

Subtract $1$ from $6$ .

$x = 5$