Solve for x log base 2 of 3x-2- log base 2 of x-5=4

$\log_{2} (3 x - 2) - \log_{2} (x - 5) = 4$

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

$\log_{2} (\frac{3 x - 2}{x - 5}) = 4$

Rewrite $\log_{2} (\frac{3 x - 2}{x - 5}) = 4$ 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 $b^{y} = x$ .

$2^{4} = \frac{3 x - 2}{x - 5}$

Cross multiply to remove the fraction.

$3 x - 2 = 2^{4} (x - 5)$

Simplify $2^{4} (x - 5)$ .

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

$3 x - 2 = 16 (x - 5)$

Apply the distributive property.

$3 x - 2 = 16 x + 16 \cdot - 5$

Multiply $16$ by $- 5$ .

$3 x - 2 = 16 x - 80$

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

Subtract $16 x$ from both sides of the equation.

$3 x - 2 - 16 x = - 80$

Subtract $16 x$ from $3 x$ .

$- 13 x - 2 = - 80$

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

Add $2$ to both sides of the equation.

$- 13 x = - 80 + 2$

Add $- 80$ and $2$ .

$- 13 x = - 78$

Divide each term in $- 13 x = - 78$ by $- 13$ and simplify.

Divide each term in $- 13 x = - 78$ by $- 13$ .

$\frac{- 13 x}{- 13} = \frac{- 78}{- 13}$

Simplify the left side.

Cancel the common factor of $- 13$ .

Cancel the common factor.

$\frac{- 13 x}{- 13} = \frac{- 78}{- 13}$

Divide $x$ by $1$ .

$x = \frac{- 78}{- 13}$

Simplify the right side.

Divide $- 78$ by $- 13$ .

$x = 6$

Solve for x log base 2 of 3x-2- log base 2 of x-5=4

Related Questions

Support LCMGCF.com by disabling your adblocker.