# Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. log2(3x−1)=log2(x+1)+3

Solve the equations and inequalities. Write the solution sets to the inequalities in interval notation. ${\mathrm{log}}_{2}\left(3x-1\right)={\mathrm{log}}_{2}\left(x+1\right)+3$

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Property used:

Property 1:

${\mathrm{log}}_{2}\left(m\right)-{\mathrm{log}}_{2}\left(n\right)={\mathrm{log}}_{2}\left(\frac{m}{n}\right)$

Now to simplifying the given equation:

${\mathrm{log}}_{2}\left(3x-1\right)={\mathrm{log}}_{2}\left(x+1\right)+3$

${\mathrm{log}}_{2}\left(3x-1\right)-{\mathrm{log}}_{2}\left(x+1\right)=3$

${\mathrm{log}}_{2}\left(\frac{3x-1}{x+1}\right)=3$ [Using Property 1.]

Now taking antilog 2 and solving:

$\left(\frac{3x-1}{x+1}\right)={2}^{3}$

$\left(3x-1\right)=8\left(x+1\right)$

$\left(3x-1\right)=8x+8$

$3x-8x=8x+1$

$-5x=9$

$x=-\frac{9}{5}$

Since,

The solution does not satisfy the given equation.

Hence there is no solution for $x\in R$