Solve for x 4- log of 3-x=3

$4 - \log (3 - x) = 3$

Move all terms not containing $\log (3 - x)$ to the right side of the equation.

Subtract $4$ from both sides of the equation.

$- \log (3 - x) = 3 - 4$

Subtract $4$ from $3$ .

$- \log (3 - x) = - 1$

Divide each term in $- \log (3 - x) = - 1$ by $- 1$ and simplify.

Divide each term in $- \log (3 - x) = - 1$ by $- 1$ .

$\frac{- \log (3 - x)}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\log (3 - x)}{1} = \frac{- 1}{- 1}$

Divide $\log (3 - x)$ by $1$ .

$\log (3 - x) = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$\log (3 - x) = 1$

Rewrite $\log (3 - x) = 1$ 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 $b^{y} = x$ .

$10^{1} = 3 - x$

Solve for $x$ .

Rewrite the equation as $3 - x = 10^{1}$ .

$3 - x = 10$

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

Subtract $3$ from both sides of the equation.

$- x = 10 - 3$

Subtract $3$ from $10$ .

$- x = 7$

Divide each term in $- x = 7$ by $- 1$ and simplify.

Divide each term in $- x = 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{7}{- 1}$

Divide $x$ by $1$ .

$x = \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x = - 7$

Solve for x 4- log of 3-x=3

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