Solve for a log base 2 of 9+ log base 2 of a = log base 2 of 11

$\log_{2} (9) + \log_{2} (a) = \log_{2} (11)$

Step 1

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{2} (9 a) = \log_{2} (11)$

Step 2

For the equation to be equal, the argument of the logarithms on both sides of the equation must be equal.

$9 a = 11$

Step 3

Divide each term in $9 a = 11$ by $9$ and simplify.

Divide each term in $9 a = 11$ by $9$ .

$\frac{9 a}{9} = \frac{11}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 a}{9} = \frac{11}{9}$

Divide $a$ by $1$ .

$a = \frac{11}{9}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$a = \frac{11}{9}$

Decimal Form:

$a = 1.\overline{2}$

Mixed Number Form:

$a = 1 \frac{2}{9}$