Solve for x log base 5 of x/5-3 = log base 5 of 7x-2

$\log_{5} (\frac{x}{5} - 3) = \log_{5} (7 x - 2)$

Step 1

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$\frac{x}{5} - 3 = 7 x - 2$

Step 2

Solve for $x$ .

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

Subtract $7 x$ from both sides of the equation.

$\frac{x}{5} - 3 - 7 x = - 2$

To write $- 7 x$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$\frac{x}{5} - 7 x \cdot \frac{5}{5} - 3 = - 2$

Combine $- 7 x$ and $\frac{5}{5}$ .

$\frac{x}{5} + \frac{- 7 x \cdot 5}{5} - 3 = - 2$

Combine the numerators over the common denominator.

$\frac{x - 7 x \cdot 5}{5} - 3 = - 2$

Simplify each term.

Simplify the numerator.

Factor $x$ out of $x - 7 x \cdot 5$ .

Raise $x$ to the power of $1$ .

$\frac{x^{1} - 7 x \cdot 5}{5} - 3 = - 2$

Factor $x$ out of $x^{1}$ .

$\frac{x \cdot 1 - 7 x \cdot 5}{5} - 3 = - 2$

Factor $x$ out of $- 7 x \cdot 5$ .

$\frac{x \cdot 1 + x (- 7 \cdot 5)}{5} - 3 = - 2$

Factor $x$ out of $x \cdot 1 + x (- 7 \cdot 5)$ .

$\frac{x (1 - 7 \cdot 5)}{5} - 3 = - 2$

Multiply $- 7$ by $5$ .

$\frac{x (1 - 35)}{5} - 3 = - 2$

Subtract $35$ from $1$ .

$\frac{x \cdot - 34}{5} - 3 = - 2$

Move $- 34$ to the left of $x$ .

$\frac{- 34 \cdot x}{5} - 3 = - 2$

Move the negative in front of the fraction.

$- \frac{34 x}{5} - 3 = - 2$

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

Add $3$ to both sides of the equation.

$- \frac{34 x}{5} = - 2 + 3$

Add $- 2$ and $3$ .

$- \frac{34 x}{5} = 1$

Multiply both sides of the equation by $- \frac{5}{34}$ .

$- \frac{5}{34} (- \frac{34 x}{5}) = - \frac{5}{34} \cdot 1$

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{5}{34} (- \frac{34 x}{5})$ .

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{5}{34}$ into the numerator.

$\frac{- 5}{34} (- \frac{34 x}{5}) = - \frac{5}{34} \cdot 1$

Move the leading negative in $- \frac{34 x}{5}$ into the numerator.

$\frac{- 5}{34} \cdot \frac{- 34 x}{5} = - \frac{5}{34} \cdot 1$

Factor $5$ out of $- 5$ .

$\frac{5 (- 1)}{34} \cdot \frac{- 34 x}{5} = - \frac{5}{34} \cdot 1$

Cancel the common factor.

$\frac{5 \cdot - 1}{34} \cdot \frac{- 34 x}{5} = - \frac{5}{34} \cdot 1$

Rewrite the expression.

$\frac{- 1}{34} (- 34 x) = - \frac{5}{34} \cdot 1$

Cancel the common factor of $34$ .

Factor $34$ out of $- 34 x$ .

$\frac{- 1}{34} (34 (- x)) = - \frac{5}{34} \cdot 1$

Cancel the common factor.

$\frac{- 1}{34} (34 (- x)) = - \frac{5}{34} \cdot 1$

Rewrite the expression.

$- - x = - \frac{5}{34} \cdot 1$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 x = - \frac{5}{34} \cdot 1$

Multiply $x$ by $1$ .

$x = - \frac{5}{34} \cdot 1$

Simplify the right side.

Multiply $- 1$ by $1$ .

$x = - \frac{5}{34}$

Step 3

Exclude the solutions that do not make $\log_{5} (\frac{x}{5} - 3) = \log_{5} (7 x - 2)$ true.

$No solution$

Solve for x log base 5 of x/5-3 = log base 5 of 7x-2

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