Solve for x log base 5 of 2x+1 = log base 5 of 4x-7

$\log_{5} (2 x) + 1 = \log_{5} (4 x) - 7$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{5} (2 x) - \log_{5} (4 x) = - 1 - 7$

Step 2

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\log_{5} (\frac{2 x}{4 x}) = - 1 - 7$

Step 3

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2 x$ .

$\log_{5} (\frac{2 (x)}{4 x}) = - 1 - 7$

Cancel the common factors.

Factor $2$ out of $4 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 4

Cancel the common factor of $x$ .

Cancel the common factor.

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

Rewrite the expression.

$\log_{5} (\frac{1}{2}) = - 1 - 7$

Step 5

Subtract $7$ from $- 1$ .

$\log_{5} (\frac{1}{2}) = - 8$

Step 6

The equation is never true.

No solution