Solve for x log base 4 of x-4=( log base 4 of x)/( log base 4 of 4)

$\log_{4} (x - 4) = \frac{\log_{4} (x)}{\log_{4} (4)}$

Simplify $\frac{\log_{4} (x)}{\log_{4} (4)}$ .

Logarithm base $4$ of $4$ is $1$ .

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

Divide $\log_{4} (x)$ by $1$ .

$\log_{4} (x - 4) = \log_{4} (x)$

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

Subtract $\log_{4} (x)$ from both sides of the equation.

$\log_{4} (x - 4) - \log_{4} (x) = 0$

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

$\log_{4} (\frac{x - 4}{x}) = 0$

Rewrite $\log_{4} (\frac{x - 4}{x}) = 0$ 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 $b^{y} = x$ .

$4^{0} = \frac{x - 4}{x}$

Cross multiply to remove the fraction.

$x - 4 = 4^{0} (x)$

Simplify $4^{0} (x)$ .

Anything raised to $0$ is $1$ .

$x - 4 = 1 x$

Multiply $x$ by $1$ .

$x - 4 = x$

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

Subtract $x$ from both sides of the equation.

$x - 4 - x = 0$

Combine the opposite terms in $x - 4 - x$ .

Subtract $x$ from $x$ .

$0 - 4 = 0$

Subtract $4$ from $0$ .

$- 4 = 0$

Add $4$ to both sides of the equation.

$0 = 4$

The equation is never true.

No solution

Solve for x log base 4 of x-4=( log base 4 of x)/( log base 4 of 4)

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