Solve for x 5+6* log base 2 of x=14

$5 + 6 \cdot \log_{2} (x) = 14$

Step 1

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

Subtract $5$ from both sides of the equation.

$6 \log_{2} (x) = 14 - 5$

Subtract $5$ from $14$ .

$6 \log_{2} (x) = 9$

Step 2

Divide each term in $6 \log_{2} (x) = 9$ by $6$ and simplify.

Divide each term in $6 \log_{2} (x) = 9$ by $6$ .

$\frac{6 \log_{2} (x)}{6} = \frac{9}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 \log_{2} (x)}{6} = \frac{9}{6}$

Divide $\log_{2} (x)$ by $1$ .

$\log_{2} (x) = \frac{9}{6}$

Simplify the right side.

Cancel the common factor of $9$ and $6$ .

Factor $3$ out of $9$ .

$\log_{2} (x) = \frac{3 (3)}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$\log_{2} (x) = \frac{3 \cdot 3}{3 \cdot 2}$

Cancel the common factor.

$\log_{2} (x) = \frac{3 \cdot 3}{3 \cdot 2}$

Rewrite the expression.

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

Step 3

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

$2^{\frac{3}{2}} = x$

Step 4

Rewrite the equation as $x = 2^{\frac{3}{2}}$ .

$x = 2^{\frac{3}{2}}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = 2^{\frac{3}{2}}$

Decimal Form:

$x = 2.82842712 \dots$

Solve for x 5+6* log base 2 of x=14

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