How do you solve ${{3}^{2x+1}}=5$?

Hint: First take logarithm on both the sides. Then try to separate the constants and the variables. Keep all the terms containing ‘x’ on the left side of the equation and all constant terms on the right side of the equation. Finally solve for ‘x’ by doing the necessary calculations to obtain the required result.

Complete step by step answer:
Solving the equation means, we have to find the value of ‘x’ for which the equation gets satisfied.
Considering our equation ${{3}^{2x+1}}=5$
Taking logarithm both the sides, we get
$\Rightarrow \log {{3}^{2x+1}}=\log 5$
As we know $\log {{m}^{n}}=n\log m$ (by logarithmic rule)
So, $\log {{3}^{2x+1}}$ can be written as $\log {{3}^{2x+1}}=\left( 2x+1 \right)\log 3$
Hence, our equation becomes
$\Rightarrow \left( 2x+1 \right)\log 3=\log 5$
Dividing both the sides by $\log 3$, we get
\[\Rightarrow \dfrac{\left( 2x+1 \right)\log 3}{\log 3}=\dfrac{\log 5}{\log 3}\]
Cancelling out \[\log 3\] from both numerator and denominator on the right side of the equation, we get
\[\Rightarrow 2x+1=\dfrac{\log 5}{\log 3}\]
Now, we have to separate the constants and variables.
Since, the term containing ‘x’ is already on the left side of the equation, so bringing all the constant terms to the right side of the equation, we get
\[\Rightarrow 2x=\dfrac{\log 5}{\log 3}-1\]
Again as we know the value of $\log 5=0.699$ and $\log 3=0.301$
So, putting the value of $\log 3$ and $\log 5$, we get
$\begin{align}
  & \Rightarrow 2x=\dfrac{0.699}{0.301}-1 \\
 & \Rightarrow 2x=2.322-1 \\
 & \Rightarrow 2x=1.322 \\
\end{align}$
Dividing ‘2’ both the sides, we get
$\Rightarrow \dfrac{2x}{2}=\dfrac{1.322}{2}$
Cancelling out 2 from both numerator and denominator on the right side of the equation, we get
$\begin{align}
  & \Rightarrow x=\dfrac{1.322}{2} \\
 & \Rightarrow x=0.661 \\
\end{align}$

Note: Some basic logarithmic rules and values should be known for maximum simplification. For example in the above calculation we encountered \[\dfrac{\log 5}{\log 3}\]. It can only be simplified when both the values of $\log 3$ and $\log 5$ are known. So, here are some basic logarithmic values:
$\begin{align}
  & \log 1=0 \\
 & \log 2=0.301 \\
 & \log 3=0.477 \\
 & \log 4=0.602 \\
 & \log 5=0.699 \\
\end{align}$