Question

In the following exercises, solve for $x$

${\log }_3(2x-1)={\log }_3(x+3)+{\log }_{3}3$

Short answer

By solving the equation ${\log }_3(2x-1)={\log }_3(x+3)+{\log }_{3}3$, the value of $x$is $-10$

Step-by-step solution

Step 1. Given

An expression${\log }_3(2x-1)={\log }_3(x+3)+{\log }_{3}3$

To solve the expression for$x$

Step 2. Use the Product property

By Product Property of logarithm,

${\log }_{a}M+{\log }_{a}N={\log }_{a}MN$

$$\begin{gathered} {\log }_3(2x-1)={\log }_3(x+3)\left(3\right) \\ {\log }_3(2x-1)={\log }_3(3x+9) \end{gathered}$$

Step 3. Use One to One property

If ${\log }_{a}M={\log }_{a}N$, then $M=N$

$$\begin{gathered} 2x-1=3x+9 \\ 3x-2x=-9-1 \\ x=-10 \end{gathered}$$

Step 5. Check the solution

Substitute $x=-10$in the original equation and solve.

$$\begin{gathered} {\log }_3\left(2\right(-10)-1)={\log }_3\left(\right(-10)+3)+{\log }_{3}3 \\ {\log }_3(-21)={\log }_3(-7)+{\log }_{3}3 \\ {\log }_3(-21)={\log }_3(-21) \end{gathered}$$

Hence the solution is checked.