Avalance789 said:
Sorry, I have confused you. Should be written like
sqrt (5-6x)*ln(4x^2-a^2)=sqrt(5-6x)*ln(2x+a)So sorry
Okay, so the equation is:
$$\sqrt{5-6x}\ln\left(4x^2-a^2\right)=\sqrt{5-6x}\ln\left(2x+a\right)$$
Let's arrange as:
$$\sqrt{5-6x}\ln\left(4x^2-a^2\right)-\sqrt{5-6x}\ln\left(2x+a\right)=0$$
Factor:
$$\sqrt{5-6x}\left(\ln\left(4x^2-a^2\right)-\ln\left(2x+a\right)\right)=0$$
Apply log rule for subtraction:
$$\sqrt{5-6x}\ln\left(\frac{4x^2-a^2}{2x+a}\right)=0$$
Factor log argument:
$$\sqrt{5-6x}\ln\left(\frac{(2x+a)(2x-a)}{2x+a}\right)=0$$
Divide out common factors:
$$\sqrt{5-6x}\ln(2x-a)=0$$
Okay, now I would first look at the square root, and observe that in order for the equation to have only 1 solution, we need:
$$5-6x>0\implies x<\frac{5}{6}$$
Now, in order for the equation to be true with respect to the factor involving \(a\), we require:
$$2x-a=1\implies x=\frac{a+1}{2}$$
And so:
$$\frac{a+1}{2}<\frac{5}{6}$$
$$a+1<\frac{5}{3}$$
$$a<\frac{2}{3}$$