Solve the Inequality for x log base 7 of 5x+3=3

$\log_{7} (5 x) + 3 = 3$

Move all terms not containing $\log_{7} (5 x)$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$\log_{7} (5 x) = 3 - 3$

Subtract $3$ from $3$ .

$\log_{7} (5 x) = 0$

Rewrite $\log_{7} (5 x) = 0$ 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 $b^{y} = x$ .

$7^{0} = 5 x$

Solve for $x$ .

Rewrite the equation as $5 x = 7^{0}$ .

$5 x = 7^{0}$

Anything raised to $0$ is $1$ .

$5 x = 1$

Divide each term in $5 x = 1$ by $5$ and simplify.

Divide each term in $5 x = 1$ by $5$ .

$\frac{5 x}{5} = \frac{1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{1}{5}$

Divide $x$ by $1$ .

$x = \frac{1}{5}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{5}$

Decimal Form:

$x = 0.2$

Solve the Inequality for x log base 7 of 5x+3=3

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