Solve for x 5+ log base 2 of -5x=9

$5 + \log_{2} (- 5 x) = 9$

Step 1

Move all terms not containing $\log_{2} (- 5 x)$ to the right side of the equation.

Subtract $5$ from both sides of the equation.

$\log_{2} (- 5 x) = 9 - 5$

Subtract $5$ from $9$ .

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

Step 2

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

$2^{4} = - 5 x$

Step 3

Solve for $x$ .

Rewrite the equation as $- 5 x = 2^{4}$ .

$- 5 x = 2^{4}$

Divide each term in $- 5 x = 2^{4}$ by $- 5$ and simplify.

Divide each term in $- 5 x = 2^{4}$ by $- 5$ .

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

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

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

$x = \frac{16}{- 5}$

Move the negative in front of the fraction.

$x = - \frac{16}{5}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{16}{5}$

Decimal Form:

$x = - 3.2$

Mixed Number Form:

$x = - 3 \frac{1}{5}$

Solve for x 5+ log base 2 of -5x=9

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