Solve for x log of 7x+1 = log of x-2+1

$\log (7 x + 1) = \log (x - 2) + 1$

Move all the terms containing a logarithm to the left side of the equation.

$\log (7 x + 1) - \log (x - 2) = 1$

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

$\log (\frac{7 x + 1}{x - 2}) = 1$

Rewrite $\log (\frac{7 x + 1}{x - 2}) = 1$ 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 $b^{y} = x$ .

$10^{1} = \frac{7 x + 1}{x - 2}$

Cross multiply to remove the fraction.

$7 x + 1 = 10 (x - 2)$

Simplify $10 (x - 2)$ .

Apply the distributive property.

$7 x + 1 = 10 x + 10 \cdot - 2$

Multiply $10$ by $- 2$ .

$7 x + 1 = 10 x - 20$

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

Subtract $10 x$ from both sides of the equation.

$7 x + 1 - 10 x = - 20$

Subtract $10 x$ from $7 x$ .

$- 3 x + 1 = - 20$

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

Subtract $1$ from both sides of the equation.

$- 3 x = - 20 - 1$

Subtract $1$ from $- 20$ .

$- 3 x = - 21$

Divide each term in $- 3 x = - 21$ by $- 3$ and simplify.

Divide each term in $- 3 x = - 21$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 21}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 21}{- 3}$

Divide $x$ by $1$ .

$x = \frac{- 21}{- 3}$

Simplify the right side.

Divide $- 21$ by $- 3$ .

$x = 7$

Solve for x log of 7x+1 = log of x-2+1

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