Solve for x log base 7 of 3-2x=3

$\log_{7} (3 - 2 x) = 3$

Step 1

Rewrite $\log_{7} (3 - 2 x) = 3$ 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 $b^{y} = x$ .

$7^{3} = 3 - 2 x$

Step 2

Solve for $x$ .

Rewrite the equation as $3 - 2 x = 7^{3}$ .

$3 - 2 x = 7^{3}$

Raise $7$ to the power of $3$ .

$3 - 2 x = 343$

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

Subtract $3$ from both sides of the equation.

$- 2 x = 343 - 3$

Subtract $3$ from $343$ .

$- 2 x = 340$

Divide each term in $- 2 x = 340$ by $- 2$ and simplify.

Divide each term in $- 2 x = 340$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{340}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{340}{- 2}$

Divide $x$ by $1$ .

$x = \frac{340}{- 2}$

Simplify the right side.

Divide $340$ by $- 2$ .

$x = - 170$

Solve for x log base 7 of 3-2x=3

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