Solve for x log of 7x+7=1

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

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

$10^{1} = 7 x + 7$

Solve for $x$ .

Rewrite the equation as $7 x + 7 = 10^{1}$ .

$7 x + 7 = 10$

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

Subtract $7$ from both sides of the equation.

$7 x = 10 - 7$

Subtract $7$ from $10$ .

$7 x = 3$

Divide each term in $7 x = 3$ by $7$ and simplify.

Divide each term in $7 x = 3$ by $7$ .

$\frac{7 x}{7} = \frac{3}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{3}{7}$

Divide $x$ by $1$ .

$x = \frac{3}{7}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{3}{7}$

Decimal Form:

$x = 0.\overline{428571}$

Solve for x log of 7x+7=1

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