What are the potential solutions of log4x+log4(x+6)=2?
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The potential solutions of [tex]log_4x+log_4(x+6)=2[/tex] are 2 and -8.
From the properties of logarithms, you can rewrite logarithmic expressions.
The main properties are:
The exercise asks the potential solutions for [tex]log_4x+log_4(x+6)=2[/tex]. In this expression you can apply the Product Rule for Logarithms.
[tex]log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0[/tex]
Now you should solve the quadratic equation.
Δ=[tex]b^2-4ac=36-4*1*(-16)=36+64=100[/tex]. Thus, x will be [tex]x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}[/tex]. Then:
[tex]x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8[/tex]
The potential solutions are 2 and -8.
Read more about the properties of logarithms here:
https://brainly.com/question/14868849