How do you solve ${5^x} = 52$?

Hint:
Though this sum looks like there is some mistake, sometimes if the power and the RHS seem to remain mismatched, it should click in the mind of the student that it is not an indices sum but a logarithm sum. In this sum, the student has to take a log on both sides and then use the properties of Log to solve the numerical. After taking the logarithm the student has to bring all the constants on one side and the variable on the other. The last step is to input the value of the log to get the final answer.

Complete Step by Step Solution:
In this particular sum we will be using the property $\log {a^b} = b\log a$
Let us take logs on both sides in the given sum. After adding log the next step is
$\log ({5^x}) = \log 52..........(1)$
Utilizing the property of Logs - $\log {a^b} = b\log a$, we have
$x\log (5) = \log 52..........(2)$
After rearranging the above equation ( getting all constants on one side and variables on other) we get the final solution
$x = \dfrac{{\log 52}}{{\log (5)}}..........(3)$
We have made no assumptions regarding the base of the logarithm, therefore using base $10$logarithms( in conjunction with the calculator)
$x = \dfrac{{{{\log }_{10}}52}}{{{{\log }_{10}}(5)}}..........(4)$
Substituting the value of $\log 52\& \log (5)$we get the final answer.

Therefore the value of $x$is $2.455$.

Note:
Whenever the student is confused about whether to use indices or Logarithm, he /she should always go for logarithm. This is because the final answer would be the same, only the difference in the number of steps would be more in the case of the logarithm approach. Students should always follow logarithm methods as it makes the sum easy to solve and without creating any complications, unlike indices where the student might forget the property of indices.