Answer:
[tex]x=\frac{4*(2+e)}{e-2}[/tex]
Step-by-step explanation:
Let's rewrite the left side keeping in mind the next propierties:
[tex]log(\frac{1}{x} )=-log(x)[/tex]
[tex]log(x*y)=log(x)+log(y)[/tex]
Therefore:
[tex]log(2*(x^{2} -16))+log(\frac{1}{(x-4)^{2} })=1\\ log(\frac{2*(x^{2} -16)}{(x-4)^{2}})=1[/tex]
Now, cancel logarithms by taking exp of both sides:
[tex]e^{log(\frac{2*(x^{2} -16)}{(x-4)^{2}})} =e^{1} \\\frac{2*(x^{2} -16)}{(x-4)^{2}}=e[/tex]
Multiply both sides by [tex](x-4)^{2}[/tex] and using distributive propierty:
[tex]2x^{2} -32=16e-8ex+ex^{2}[/tex]
Substract [tex]16e-8ex+ex^{2}[/tex] from both sides and factoring:
[tex]-(x-4)*(-8-4e-2x+ex)=0[/tex]
Multiply both sides by -1:
[tex](x-4)*(-8-4e-2x+ex)=0[/tex]
Split into two equations:
[tex]x-4=0\hspace{3}or\hspace{3}-8-4e-2x+ex=0[/tex]
Solving for [tex]x-4=0[/tex]
Add 4 to both sides:
[tex]x=4[/tex]
Solving for [tex]-8-4e-2x+ex=0[/tex]
Collect in terms of x and add [tex]4e+8[/tex] to both sides:
[tex]x(e-2)=4e+8[/tex]
Divide both sides by e-2:
[tex]x=\frac{4*(2+e)}{e-2}[/tex]
The solutions are:
[tex]x=4\hspace{3}or\hspace{3}x=\frac{4*(2+e)}{e-2}[/tex]
If we evaluate x=4 in the original equation:
[tex]log(0)-log(0)=1[/tex]
This is an absurd because log (x) is undefined for [tex]x\leq 0[/tex]
If we evaluate [tex]x=\frac{4*(2+e)}{e-2}[/tex] in the original equation:
[tex]log(2*((\frac{4e+8}{e-2})^2-16))-log((\frac{4e+8}{e-2}-4)^2)=1[/tex]
Which is correct, therefore the solution is:
[tex]x=\frac{4*(2+e)}{e-2}[/tex]
