Evaluate log base 4 of x+ log base 4 of 3x-1=1

$\log_{4} (x) + \log_{4} (3 x - 1) = 1$

Step 1

Simplify the left side.

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

$\log_{4} (x (3 x - 1)) = 1$

Simplify by multiplying through.

Apply the distributive property.

$\log_{4} (x (3 x) + x \cdot - 1) = 1$

Reorder.

Rewrite using the commutative property of multiplication.

$\log_{4} (3 x \cdot x + x \cdot - 1) = 1$

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

$\log_{4} (3 x \cdot x - 1 \cdot x) = 1$

Simplify each term.

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

Move $x$ .

$\log_{4} (3 (x \cdot x) - 1 \cdot x) = 1$

Multiply $x$ by $x$ .

$\log_{4} (3 x^{2} - 1 \cdot x) = 1$

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

$\log_{4} (3 x^{2} - x) = 1$

Step 2

Rewrite $\log_{4} (3 x^{2} - x) = 1$ 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 $b^{y} = x$ .

$4^{1} = 3 x^{2} - x$

Step 3

Solve for $x$ .

Rewrite the equation as $3 x^{2} - x = 4^{1}$ .

$3 x^{2} - x = 4$

Subtract $4$ from both sides of the equation.

$3 x^{2} - x - 4 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 4 = - 12$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$3 x^{2} - (x) - 4 = 0$

Rewrite $- 1$ as $3$ plus $- 4$

$3 x^{2} + (3 - 4) x - 4 = 0$

Apply the distributive property.

$3 x^{2} + 3 x - 4 x - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} + 3 x) - 4 x - 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$3 x (x + 1) - 4 (x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 1$ .

$(x + 1) (3 x - 4) = 0$

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

$x + 1 = 0$

$3 x - 4 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $3 x - 4$ equal to $0$ and solve for $x$ .

Set $3 x - 4$ equal to $0$ .

$3 x - 4 = 0$

Solve $3 x - 4 = 0$ for $x$ .

Add $4$ to both sides of the equation.

$3 x = 4$

Divide each term in $3 x = 4$ by $3$ and simplify.

Divide each term in $3 x = 4$ by $3$ .

$\frac{3 x}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{4}{3}$

Divide $x$ by $1$ .

$x = \frac{4}{3}$

The final solution is all the values that make $(x + 1) (3 x - 4) = 0$ true.

$x = - 1, \frac{4}{3}$

Step 4

Exclude the solutions that do not make $\log_{4} (x) + \log_{4} (3 x - 1) = 1$ true.

$x = \frac{4}{3}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \frac{4}{3}$

Decimal Form:

$x = 1.\overline{3}$

Mixed Number Form:

$x = 1 \frac{1}{3}$

Evaluate log base 4 of x+ log base 4 of 3x-1=1

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