How do you solve \[{{3}^{5x}}={{27}^{2x+1}}\]?

Hint: To solve this question, we should know some of the logarithmic properties. We will use the property that \[\log {{a}^{b}}=b\log a\]. We should also know that to solve a linear equation, we have to take all the variable terms to one side of the equation, and the constant terms to the other sides of the equation. By this, we can find the solution value of the variable.

Complete step by step solution:
We are given the equation \[{{3}^{5x}}={{27}^{2x+1}}\]. Taking logarithm to both sides of the equation, we get \[\log {{3}^{5x}}=\log {{27}^{2x+1}}\]. Using the logarithmic property \[\log {{a}^{b}}=b\log a\] in this equation, we get
\[\Rightarrow 5x\log 3=\left( 2x+1 \right)\log 27\]
We know that 27 is the cube of 2, hence the equation can also be written as
\[\Rightarrow 5x\log 3=\left( 2x+1 \right)\log {{3}^{3}}\]
Again, using the property \[\log {{a}^{b}}=b\log a\] in the RHS of the above equation, we get
\[\Rightarrow 5x\log 3=3\left( 2x+1 \right)\log 3\]
canceling out common factor form both sides of the equation, we get
\[\Rightarrow 5x=3\left( 2x+1 \right)\]
As we can see this is a linear equation, we know that to solve a linear equation, we need to take the variable terms to one side of the equation. Simplifying the above equation, we get
\[\Rightarrow 5x=6x+3\]
Subtracting \[6x\] from both sides of the above equation, we get
\[\begin{align}
  & \Rightarrow 5x=3\left( 2x+1 \right) \\
 & \Rightarrow -x=3 \\
 & \Rightarrow x=-3 \\
\end{align}\]
Thus, the solution for the equation is \[x=-3\].

Note: We can check if the solution is correct or not by substituting the value in the equation. The LHS of the equation is \[{{3}^{5x}}\], and the RHS is \[{{27}^{2x+1}}\]. substituting \[x=-3\] in both sides of the equation, we get LHS as \[{{3}^{5(-3)}}={{3}^{-15}}\], and RHS as \[{{27}^{2(-3)+1}}={{27}^{-5}}\], as 27 is the cube of 3 \[{{27}^{-5}}={{3}^{3\left( -5 \right)}}={{3}^{-15}}\]. As LHS and RHS are the same, the solution is correct.