Write in Standard Form log base 7 of x+ log base 7 of 6x-1=1

$\log_{7} (x) + \log_{7} (6 x - 1) = 1$

Step 1

Subtract $1$ from both sides of the equation.

$\log_{7} (x) + \log_{7} (6 x - 1) - 1 = 0$

Step 2

Simplify $\log_{7} (x) + \log_{7} (6 x - 1) - 1$ .

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{7} (x (6 x - 1)) - 1 = 0$

Simplify each term.

Apply the distributive property.

$\log_{7} (x (6 x) + x \cdot - 1) - 1 = 0$

Rewrite using the commutative property of multiplication.

$\log_{7} (6 x \cdot x + x \cdot - 1) - 1 = 0$

Move $- 1$ to the left of $x$ .

$\log_{7} (6 x \cdot x - 1 \cdot x) - 1 = 0$

Simplify each term.

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$\log_{7} (6 (x \cdot x) - 1 \cdot x) - 1 = 0$

Multiply $x$ by $x$ .

$\log_{7} (6 x^{2} - 1 \cdot x) - 1 = 0$

Rewrite $- 1 x$ as $- x$ .

$\log_{7} (6 x^{2} - x) - 1 = 0$

Step 3

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$a x^{2} + b x + c = 0$

Step 4

The standard form is $\log_{7} (6 x^{2} - x) - 1$ .

$\log_{7} (6 x^{2} - x) - 1 = 0$

Step 5