How do you solve ${{6}^{x-2}}={{4}^{x}}$ ?

Hint: To solve exponential function, there are numerous properties of exponents that are suitable if the bases are the same or could be reduced to the same base. But if bases are neither the same nor could be written as a power of the same base, then we can use the logarithm function to solve these equations. To further simplify this, we use all the properties of the logarithm. In this type of problem, we use the common or natural logarithmic function.

Complete Step by Step Solution:
We have to solve the equation ${{6}^{x-2}}={{4}^{x}}$
That is we have to find the value of $x$ such that it satisfies the equation.
We can do this by taking log of both the sides because $6$ cannot be written as the power of $4$
( Also , we avoid taking log of 0 )
$\log {{6}^{x-2}}=\log {{4}^{x}}$
Now using the property
$log{{a}^{b}}=bloga$
We can write
$(x-2)log6=xlog4$
$xlog6-2log6=xlog4$
$xlog6-xlog4=2log6$
$x(log6-log4)=2log6$
$x=\dfrac{2log6}{(log6-log4)}$
$x=\dfrac{2log6}{(log(2\times 3)-log({{2}^{2}}))}$
Using the property \[lo{{g}_{a}}\text{ }\left( x\times y \right)=lo{{g}_{a}}\text{ }x+lo{{g}_{a}}\text{ }y\]
We get
\[\begin{array}{*{35}{l}}
   x=\dfrac{2log6}{\left( log2+log\text{ }3-2log2 \right)} \\
  x=\dfrac{2log6}{\left( log3-log2 \right)} \\
\end{array}\]

Note:
We usually avoid taking the logarithm of 0 because it is undefined. We can also change the base of the logarithmic functions as well as we can switch the base and exponential using some specific rules. However, the change of bases is possible if both the base and the number are greater than $0$.