if the equation below is solved by graphing, which statement is true?\nlog(6x + 10) = log_{\\frac{1}{2}}x\nth…

Explanation:

Step1: Analyze the left - hand side function (y = \log(6x + 10))

The domain of (y=\log(6x + 10)) is (6x+10>0), i.e., (x>-\frac{10}{6}\approx - 1.67).

Step2: Analyze the right - hand side function (y=\log_{\frac{1}{2}}x)

The domain of (y = \log_{\frac{1}{2}}x) is (x>0).

Step3: Use a graphing utility

When we graph (y=\log(6x + 10)) (where (y=\frac{\ln(6x + 10)}{\ln10})) and (y=\log_{\frac{1}{2}}x=\frac{\ln x}{\ln\frac{1}{2}}=-\frac{\ln x}{\ln2}) using a graphing calculator or software:

• Substitute (x = 0.46):
  • For (y_1=\log(6x + 10)), (y_1=\log(6\times0.46+10)=\log(2.76 + 10)=\log(12.76)\approx1.106)
  • For (y_2=\log_{\frac{1}{2}}x), (y_2=\log_{\frac{1}{2}}0.46=-\frac{\ln0.46}{\ln2}\approx1.12)
• Substitute (x = 0.75):
  • (y_1=\log(6\times0.75+10)=\log(4.5 + 10)=\log(14.5)\approx1.16)
  • (y_2=\log_{\frac{1}{2}}0.75=-\frac{\ln0.75}{\ln2}\approx0.415)
• Substitute (x = 1.11):
  • (y_1=\log(6\times1.11+10)=\log(6.66+10)=\log(16.66)\approx1.22)
  • (y_2=\log_{\frac{1}{2}}1.11=-\frac{\ln1.11}{\ln2}\approx - 0.13)
• Substitute (x = 3.07):
  • (y_1=\log(6\times3.07+10)=\log(18.42+10)=\log(28.42)\approx1.45)
  • (y_2=\log_{\frac{1}{2}}3.07=-\frac{\ln3.07}{\ln2}\approx - 1.62)

Answer:

The curves intersect at approximately (x = 0.46).