Set y = 0 in the equation y = lo g 3 ( x + 2 ) − 1 .
Solve for x in the equation 0 = lo g 3 ( x + 2 ) − 1 .
Rewrite the equation in exponential form: 3 1 = x + 2 .
Solve for x : x = 3 − 2 = 1 , so the point is ( 1 , 0 ) .
Explanation
Understanding the Problem We are given the equation y = lo g 3 ( x + 2 ) − 1 and asked to find the point that corresponds to the real zero of the graph. This means we need to find the point ( x , y ) on the graph where y = 0 .
Setting y = 0 To find the real zero, we set y = 0 in the equation: 0 = lo g 3 ( x + 2 ) − 1
Isolating the Logarithm Add 1 to both sides of the equation: 1 = lo g 3 ( x + 2 )
Converting to Exponential Form Rewrite the equation in exponential form: 3 1 = x + 2
Solving for x Solve for x : x = 3 − 2 = 1
Finding the Point So, the real zero occurs at x = 1 . This means the point on the graph where y = 0 is ( 1 , 0 ) .
Checking the Options We check the given options to see if ( 1 , 0 ) is among them. Indeed, ( 1 , 0 ) is one of the options.
Final Answer Therefore, the point that corresponds to the real zero of the graph of y = lo g 3 ( x + 2 ) − 1 is ( 1 , 0 ) .
Examples
Logarithmic functions are used in various real-world applications, such as measuring the intensity of earthquakes on the Richter scale. The Richter scale uses a logarithmic scale to quantify the magnitude of an earthquake based on the amplitude of seismic waves recorded on seismographs. Similarly, in chemistry, pH values, which indicate the acidity or alkalinity of a solution, are measured using a logarithmic scale. Understanding logarithmic functions helps scientists and engineers analyze and interpret data in these fields.
The real zero of the graph y = lo g 3 ( x + 2 ) − 1 is found by setting y = 0 . Solution leads to the point ( 1 , 0 ) , which is option B among the provided choices. Therefore, the correct answer is ( 1 , 0 ) .
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