Solve for x log base x of 7x=2

$\log_{x} (7 x) = 2$

Rewrite $\log_{x} (7 x) = 2$ 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 $b^{y} = x$ .

$x^{2} = 7 x$

Solve for $x$ .

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

$x^{2} - 7 x = 0$

Factor $x$ out of $x^{2} - 7 x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x - 7 x = 0$

Factor $x$ out of $- 7 x$ .

$x \cdot x + x \cdot - 7 = 0$

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

$x (x - 7) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$x - 7 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Add $7$ to both sides of the equation.

$x = 7$

The final solution is all the values that make $x (x - 7) = 0$ true.

$x = 0, 7$

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

$x = 7$

Solve for x log base x of 7x=2

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