Question

How do you solve ${{\log }_{3}}\left( 2x+1 \right)=2$ ? \[\]

Answer 1

Hint: We recall the definition of logarithm with base $b$ and argument $x$ as ${{b}^{y}}=x\Leftrightarrow {{\log }_{b}}x=y$. Here we are given the logarithmic equation ${{\log }_{3}}\left( 2x+1 \right)=2$ argument is $\left( 2x+1 \right)$ and base is 3 . We use the definition of logarithm and obtain a linear equation which we solve. \[\]

Complete step by step answer:
We know that the logarithm is the inverse operation to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$ must be raised, to produce that number$x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
\[{{\log }_{b}}x=y\]
Here $x$ is called the argument of the logarithm. The argument of the logarithm of the logarithm is always positive $\left( x > 0 \right)$ and the base is also positive and never equal to 1 $\left( b>0,b\ne 1 \right)$. We are give the following logarithmic equation with unknown $x$ as
\[{{\log }_{3}}\left( 2x+1 \right)=2\]
We use the definition of logarithm for base $b=3$, argument $x=2x+1$ and value of logarithm $y=2$ in exponential form to have
\[\begin{align}
  & \Rightarrow {{3}^{2}}=2x+1 \\
 & \Rightarrow 9=2x+1 \\
\end{align}\]
We subtract both sides of the above equation by 1 to have
\[\begin{align}
  & \Rightarrow 9-1=2x+1-1 \\
 & \Rightarrow 8=2x \\
\end{align}\]
We divide both sides of the above equation by 2 to have
\[\Rightarrow x=\dfrac{8}{2}=4\]

So the solution of the given equation is $x=4$. \[\]

Note: We put the value in the left hand side to have
\[\begin{align}
  & \Rightarrow {{\log }_{3}}\left( 2x+1 \right)={{\log }_{3}}\left( 2\times 4+1 \right) \\
 & \Rightarrow {{\log }_{3}}\left( 2x+1 \right){{\log }_{3}}9 \\
 & \Rightarrow {{\log }_{3}}\left( 2x+1 \right)={{\log }_{3}}{{3}^{2}} \\
\end{align}\]
We use the logarithmic identify of power ${{\log }_{b}}{{x}^{m}}=m{{\log }_{b}}x,m\ne 0$ for $x=b=3,m=2$ and identity where base and argument are equal we have ${{\log }_{b}}b=1$ for $x=b=3$ in the above step to have
\[\begin{align}
  & \Rightarrow {{\log }_{3}}\left( 2x+1 \right)=2{{\log }_{3}}3 \\
 & \Rightarrow {{\log }_{3}}\left( 2x+1 \right)=2\times 1=2 \\
\end{align}\]
Since the argument is always positive here we have $ 2x+1 > 0\Rightarrow x > \dfrac{-1}{2}$. So our solution $x=4$ is valid. We can alternatively using the identity ${{b}^{{{\log }_{b}}x}}=x$ to raise both sides of the given equation of the power 3 and then solve $2x+1=9$.