Solve for x log base 4 of 4x+2 log base 4 of x=4

$\log_{4} (4 x) + 2 \log_{4} (x) = 4$

Step 1

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

$\log_{4} (4 x) + 2 \log_{4} (x) = 4$

Step 2

Simplify the left side.

Simplify $\log_{4} (4 x) + 2 \log_{4} (x)$ .

Simplify $2 \log_{4} (x)$ by moving $2$ inside the logarithm.

$\log_{4} (4 x) + \log_{4} (x^{2}) = 4$

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

$\log_{4} (4 x \cdot x^{2}) = 4$

Multiply $x$ by $x^{2}$ by adding the exponents.

Move $x^{2}$ .

$\log_{4} (4 (x^{2} x)) = 4$

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

$\log_{4} (4 (x^{2} x^{1})) = 4$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\log_{4} (4 x^{2 + 1}) = 4$

Add $2$ and $1$ .

$\log_{4} (4 x^{3}) = 4$

Step 3

Rewrite $\log_{4} (4 x^{3}) = 4$ 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 $b^{y} = x$ .

$4^{4} = 4 x^{3}$

Step 4

Solve for $x$ .

Rewrite the equation as $4 x^{3} = 4^{4}$ .

$4 x^{3} = 4^{4}$

Subtract $4^{4}$ from both sides of the equation.

$4 x^{3} - 4^{4} = 0$

Simplify each term.

Raise $4$ to the power of $4$ .

$4 x^{3} - 1 \cdot 256 = 0$

Multiply $- 1$ by $256$ .

$4 x^{3} - 256 = 0$

Factor the left side of the equation.

Factor $4$ out of $4 x^{3} - 256$ .

Factor $4$ out of $4 x^{3}$ .

$4 (x^{3}) - 256 = 0$

Factor $4$ out of $- 256$ .

$4 x^{3} + 4 \cdot - 64 = 0$

Factor $4$ out of $4 x^{3} + 4 \cdot - 64$ .

$4 (x^{3} - 64) = 0$

Rewrite $64$ as $4^{3}$ .

$4 (x^{3} - 4^{3}) = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 4$ .

$4 ((x - 4) (x^{2} + x \cdot 4 + 4^{2})) = 0$

Factor.

Simplify.

Move $4$ to the left of $x$ .

$4 ((x - 4) (x^{2} + 4 x + 4^{2})) = 0$

Raise $4$ to the power of $2$ .

$4 ((x - 4) (x^{2} + 4 x + 16)) = 0$

Remove unnecessary parentheses.

$4 (x - 4) (x^{2} + 4 x + 16) = 0$

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

$x - 4 = 0$

$x^{2} + 4 x + 16 = 0$

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 $x^{2} + 4 x + 16$ equal to $0$ and solve for $x$ .

Set $x^{2} + 4 x + 16$ equal to $0$ .

$x^{2} + 4 x + 16 = 0$

Solve $x^{2} + 4 x + 16 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 4$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 16$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 4 \pm \sqrt{16 - 4 \cdot 16}}{2 \cdot 1}$

Multiply $- 4$ by $16$ .

$x = \frac{- 4 \pm \sqrt{16 - 64}}{2 \cdot 1}$

Subtract $64$ from $16$ .

$x = \frac{- 4 \pm \sqrt{- 48}}{2 \cdot 1}$

Rewrite $- 48$ as $- 1 (48)$ .

$x = \frac{- 4 \pm \sqrt{- 1 \cdot 48}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (48)}$ as $\sqrt{- 1} \cdot \sqrt{48}$ .

$x = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{48}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 4 \pm i \cdot \sqrt{48}}{2 \cdot 1}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$x = \frac{- 4 \pm i \cdot \sqrt{16 (3)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$x = \frac{- 4 \pm i \cdot \sqrt{4^{2} \cdot 3}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 4 \pm i \cdot (4 \sqrt{3})}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$x = \frac{- 4 \pm 4 i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 4 \pm 4 i \sqrt{3}}{2}$

Simplify $\frac{- 4 \pm 4 i \sqrt{3}}{2}$ .

$x = - 2 \pm 2 i \sqrt{3}$

The final answer is the combination of both solutions.

$x = - 2 + 2 i \sqrt{3}, - 2 - 2 i \sqrt{3}$

The final solution is all the values that make $4 (x - 4) (x^{2} + 4 x + 16) = 0$ true.

$x = 4, - 2 + 2 i \sqrt{3}, - 2 - 2 i \sqrt{3}$

Solve for x log base 4 of 4x+2 log base 4 of x=4

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