91Ó°ÊÓ

Given equation:log₃(x + 1) + log₃(2x - 3) = log₉ 9

We want to solve given equation for the value of x.

Logarithmic function is good defined only if the argument of logarithmic function is a positive real number (log_b x, x > 0).

Therefore, a given logarithmic function is good defined if the arguments:

• ${\log }_3(x+1):x+1>0⇒x>-1$
  
• ${\log }_3(2x-3):2x-3>0⇒x>\frac{3}{2}$
  

role="math" localid="1647416270134" $x∈(-1,∞)âˆ\copyright (\frac{3}{2},+∞)⇒x∈(\frac{3}{2},+∞)$

The solution of this equation must be in the interval $x∈(\frac{3}{2},+∞)---(*)$

We will use the following property for logarithmic functions:

$$\begin{gathered} r={\log }_a{a}^r(Inourcase1={\log }_3{3}^1={\log }_{3}3)---\left(1\right) \\ {\log }_{b}M+{\log }_{b}N={\log }_b(M·N)---\left(2\right) \\ If{\log }_{a}M={\log }_{a}N,thenM=N(whereM,Na>0,a≠1)---\left(3\right) \end{gathered}$$

So we have

$$\begin{gathered} {\log }_3(x+1)+{\log }_3(2x-3)={\log }_{9}9 \\ U\sin gProperty\left(1\right)and\left(2\right): \\ {\log }_3(x+1)·(2x-3)=1 \\ {\log }_3(x+1)·(2x-3)={\log }_{3}3 \\ U\sin gproperty\left(3\right): \\ (x+1)·(2x-3)=3 \\ 2x^2-3x+2x-3=3 \\ 2x^2-x-6=0 \end{gathered}$$

Notice that the bases are not equal but using the property: log_a a = 1 we get log₉ 9 = 1

We will solve this quadratic equation using formula:

$x_{1,2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$ where a, b, and c are coefficients of equation ax²+bx+c=0

So we have

$$\begin{gathered} x_{1,2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ {x}_{1,2}=\frac{1Â\pm \sqrt{1-4\left(2\right)(-6)}}{2\left(2\right)} \\ {x}_{1,2}=\frac{1Â\pm \sqrt{49}}{4} \\ {x}_{1,2}=\frac{1Â\pm 7}{4} \\ {x}_1=\frac{1-7}{4},x_2=\frac{1+7}{4} \\ {x}_1=-\frac{3}{2},x_2=2 \end{gathered}$$

Because the solution of the equation must be within the interval $x∈(\frac{3}{2},+∞)$(because of (*)) the first solution is not the solution of the given equation.

The solution of the equation is only x₂ = 2 .