Question

What are the potential solutions of log4x+log4(x+6)=2?

Answer 1

The potential solutions of $log_4x+log_4(x+6)=2$ are 2 and -8.

Properties of Logarithms

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

• Product Rule for Logarithms - $log_{b}(a*c)=log_{b}a+log_{b}c$
• Quotient Rule for Logarithms - $log_{b}(\frac{a}{c} )=log_{b}a-log_{b}c$
• Power Rule for Logarithms - $log_{b}(a^c)=c*log_{b}a$

The exercise asks the potential solutions for  $log_4x+log_4(x+6)=2$. In this expression you can apply the Product Rule for Logarithms.

                                  $log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0$

Now you should solve the quadratic equation.

 

 Δ=$b^2-4ac=36-4*1*(-16)=36+64=100$. Thus, x will be $x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}$. Then:

$x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8$

The potential solutions  are 2 and -8.

Read more about the properties of logarithms here:

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