Find the Roots (Zeros) f(x) = log base 2 of x+3

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

Step 1

Set $\log_{2} (x + 3)$ equal to $0$ .

$\log_{2} (x + 3) = 0$

Step 2

Solve for $x$ .

Rewrite $\log_{2} (x + 3) = 0$ 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 $b^{y} = x$ .

$2^{0} = x + 3$

Solve for $x$ .

Rewrite the equation as $x + 3 = 2^{0}$ .

$x + 3 = 2^{0}$

Anything raised to $0$ is $1$ .

$x + 3 = 1$

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

Subtract $3$ from both sides of the equation.

$x = 1 - 3$

Subtract $3$ from $1$ .

$x = - 2$

Step 3

Find the Roots (Zeros) f(x) = log base 2 of x+3

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