How to solve this logarithm equation? $$\frac12\cdot[\log(x) + \log(2)) + \log[\sqrt{2x} + 1] = \log(6).$$
The answer is $2$.
I've tried to solve it, but I don't know how to proceed:
$\frac12\log(2x) + \log[\sqrt{2x} + 1] = \log(6)$
$\log([(2x)^\frac12] + \log[\sqrt{2x} + 1] = \log(6)$
$\log[\sqrt{2x}] + \log[\sqrt{2x} + 1] = \log(6)$
$\log(\sqrt{2x}\cdot [\sqrt{2x} + 1]) = \log(6)$
$\sqrt{2x}\cdot[\sqrt{2x} + 1] = 6$
$2x + \sqrt{2x} = 6$
$2x + \sqrt{2}\sqrt{x} - 6 = 0$
On
$$\frac 12 [\log(x)+\log(2)]+\log(\sqrt{2x}+1)=\log(6)$$ By $\log(a^n)=n\log(a)$, we get: $$\log(\sqrt x)+\log(\sqrt 2)+\log(\sqrt{2x}+1)=\log(6)$$ By $\log(p)+\log(q)=\log(pq)$, we get: $$\log(\sqrt x\cdot\sqrt 2\cdot(\sqrt{2x}+1))=\log(6)$$ $$\log(2x+\sqrt{2x})=\log(6)$$
$$\to 2x+\sqrt{2x}=6$$
One can then iterate: $$x_{n+1}=\frac 12(6-\sqrt{2x_n}); x_0=1$$ and show this converges to $x=2$, testing this value shows it as correct.
Square both sides of the equation $\sqrt{2x}=6-2x$ to obtain $2x=36-24x+4x^2$ which will give you this quadratic equation $4x^2-26x+36=0$. If we simplify the equation further by dividing by the common factor $2$ we obtain $2x^2-13x+18=0$. This should be easy to solve if you use the quadratic formula or by factorizing the left side and finding its roots.