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

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

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

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

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

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

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

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

Cross multiply to remove the fraction.

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

Simplify $2^{4} (3 x + 5)$ .

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

$x = 16 (3 x + 5)$

Apply the distributive property.

$x = 16 (3 x) + 16 \cdot 5$

Multiply.

Multiply $3$ by $16$ .

$x = 48 x + 16 \cdot 5$

Multiply $16$ by $5$ .

$x = 48 x + 80$

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

Subtract $48 x$ from both sides of the equation.

$x - 48 x = 80$

Subtract $48 x$ from $x$ .

$- 47 x = 80$

Divide each term in $- 47 x = 80$ by $- 47$ and simplify.

Divide each term in $- 47 x = 80$ by $- 47$ .

$\frac{- 47 x}{- 47} = \frac{80}{- 47}$

Simplify the left side.

Cancel the common factor of $- 47$ .

Cancel the common factor.

$\frac{- 47 x}{- 47} = \frac{80}{- 47}$

Divide $x$ by $1$ .

$x = \frac{80}{- 47}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{80}{47}$

Exclude the solutions that do not make $\log_{2} (x) = \log_{2} (3 x + 5) + 4$ true.

$No solution$

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

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