Solve for x log base 4 of 3x-2- log base 4 of 4x+1=2

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

Step 1

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

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

Step 2

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

$4^{2} = \frac{3 x - 2}{4 x + 1}$

Step 3

Cross multiply to remove the fraction.

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

Step 4

Simplify $4^{2} (4 x + 1)$ .

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

$3 x - 2 = 16 (4 x + 1)$

Apply the distributive property.

$3 x - 2 = 16 (4 x) + 16 \cdot 1$

Multiply.

Multiply $4$ by $16$ .

$3 x - 2 = 64 x + 16 \cdot 1$

Multiply $16$ by $1$ .

$3 x - 2 = 64 x + 16$

Step 5

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

Subtract $64 x$ from both sides of the equation.

$3 x - 2 - 64 x = 16$

Subtract $64 x$ from $3 x$ .

$- 61 x - 2 = 16$

Step 6

Move all terms not containing $x$ to the right side of the equation.

Add $2$ to both sides of the equation.

$- 61 x = 16 + 2$

Add $16$ and $2$ .

$- 61 x = 18$

Step 7

Divide each term in $- 61 x = 18$ by $- 61$ and simplify.

Divide each term in $- 61 x = 18$ by $- 61$ .

$\frac{- 61 x}{- 61} = \frac{18}{- 61}$

Simplify the left side.

Cancel the common factor of $- 61$ .

Cancel the common factor.

$\frac{- 61 x}{- 61} = \frac{18}{- 61}$

Divide $x$ by $1$ .

$x = \frac{18}{- 61}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{18}{61}$

Step 8

Exclude the solutions that do not make $\log_{4} (3 x - 2) - \log_{4} (4 x + 1) = 2$ true.

$No solution$

Solve for x log base 4 of 3x-2- log base 4 of 4x+1=2

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