Question

Determine the \(x\) -intercept of \(y=\log _{7}(x+2)\).

Short answer

The x-intercept is at \( x = -1 \).

Step-by-step solution

Understand the x-intercept

The x-intercept is the point where the graph of the equation crosses the x-axis. At the x-intercept, the value of y is 0.

Set y to 0 in the equation

Given the equation:\[ y = \log _{7}(x+2) \]. To find the x-intercept, set y to 0:\[ 0 = \log _{7}(x+2) \].

Solve the logarithmic equation

To solve \log _{7}(x+2) = 0, rewrite the equation in exponential form: \( 7^0 = x+2 \).

Simplify the exponential equation

Since any number raised to the power of 0 is 1, we get: \( 1 = x + 2 \).

Solve for x

Subtract 2 from both sides of the equation to find x: \( x = 1 - 2 \), which simplifies to \( x = -1 \).

Verify the solution

Substitute \( x = -1\) back into the original equation to ensure it satisfies \( y = \log _{7}(x+2) \): \( y = \log _{7}((-1)+2) = \log _{7}(1) = 0 \), confirming that the solution is correct.

Key concepts

Understanding x-intercepts

To understand the x-intercept of a function, you need to remember that it is the point where the graph touches or crosses the x-axis. At this point, the value of the function, which is y, is always zero.
In simpler terms, for any equation, if we set y to zero, we can find where the equation intersects the x-axis. For example, in the equation given in the problem, we set y to 0 to find the x-intercept of the function.

Solving logarithmic equations

Logarithmic equations may seem complicated at first, but they're manageable if you follow each step carefully. First, understand the properties of logarithms. A logarithm answers the question: 'To what exponent must we raise a certain base to obtain a given number?'.
Let's take the exercise's equation: \( y = \log _{7}(x+2) \). To find the x-intercept, we set y to 0. This gives us \( 0 = \log _{7}(x+2) \). This means the exponent that makes the base 7 equal to \( x+2 \) is zero. To solve this, we rewrite it in exponential form.

Rewriting in exponential form

Rewriting logarithmic equations in exponential form can simplify solving them. The rule here is: if \( \log _{b}(a) = c \), this is equivalent to saying \( b^c = a \).
In our case, \( \log _{7}(x+2) = 0 \), that's equivalent to \( 7^0 = x+2 \). We know any number to the power of 0 is 1, so we get \( 1 = x+2 \). Now, it simplifies to solving the simple linear equation: Subtract 2 from both sides to get \( x = -1 \). This process simplifies solving logarithmic functions and makes equations easier to handle.