Solve for x log base x of 5x-4=2

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

Step 1

Rewrite $\log_{x} (5 x - 4) = 2$ 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 $b^{y} = x$ .

$x^{2} = 5 x - 4$

Step 2

Solve for $x$ .

Subtract $5 x$ from both sides of the equation.

$x^{2} - 5 x = - 4$

Add $4$ to both sides of the equation.

$x^{2} - 5 x + 4 = 0$

Factor $x^{2} - 5 x + 4$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $4$ and whose sum is $- 5$ .

$- 4, - 1$

Write the factored form using these integers.

$(x - 4) (x - 1) = 0$

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

$x - 4 = 0$

$x - 1 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 4) (x - 1) = 0$ true.

$x = 4, 1$

Step 3

Exclude the solutions that do not make $\log_{x} (5 x - 4) = 2$ true.

$x = 4$

Solve for x log base x of 5x-4=2

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