Question

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

Answer 1

we are given with the logarithmic equation

$log4x+log4(x+6)=2$

Using logarithmic Property

$Loga+Logb=Logab$

we can write

$Log[4x*4(x+6)]=2\\ \\ Log[16(x^2+6x)]=2$

Take anti-logarithm we get

$16(x^2+6x)=10^2\\ \\ 16x^2+96x=100\\ \\ 4x^2+24x=25\\ \\ 4x^2+24x-25=0$

Now using Quadratic formula , we can find x as below:

$x=\frac{-24 \pm\sqrt{24^2-4*4*(-25)}}{2*(4)}\\ \\ x= \frac{1}{8}(-24\pm4\sqrt{61})$

As we know the logarithm is defined for the negative values.

Hence the possible solution is

$x= \frac{1}{8}(-24+4\sqrt{61})\\ \\ x=0.905$