Find the x and y Intercepts y = log base 2 of 2x+3

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Algebra

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

Find the x-intercepts.

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To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

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

Solve the equation.

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Rewrite the equation as $\log_{2} (2 x + 3) = 0$ .

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

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

$2^{0} = 2 x + 3$

Solve for $x$ .

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

$2 x + 3 = 2^{0}$

Anything raised to $0$ is $1$ .

$2 x + 3 = 1$

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

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Subtract $3$ from both sides of the equation.

$2 x = 1 - 3$

Subtract $3$ from $1$ .

$2 x = - 2$

Divide each term in $2 x = - 2$ by $2$ and simplify.

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Divide each term in $2 x = - 2$ by $2$ .

$\frac{2 x}{2} = \frac{- 2}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{- 2}{2}$

Divide $x$ by $1$ .

$x = \frac{- 2}{2}$

Simplify the right side.

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Divide $- 2$ by $2$ .

$x = - 1$

x-intercept(s) in point form.

x-intercept(s): $(- 1, 0)$

Find the y-intercepts.

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To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

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

Solve the equation.

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Remove parentheses.

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

Simplify $\log_{2} (2 (0) + 3)$ .

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Multiply $2$ by $0$ .

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

Add $0$ and $3$ .

$y = \log_{2} (3)$

y-intercept(s) in point form.

y-intercept(s): $(0, \log_{2} (3))$

List the intersections.

x-intercept(s): $(- 1, 0)$

y-intercept(s): $(0, \log_{2} (3))$

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