Question

How do you solve \[{{5}^{x}}+1=625\]?

Answer 1

Hint: From the question given, we have been asked to solve \[{{5}^{x}}+1=625\]. We can solve the above-given equation from the question given by using some simple transformations. After using the transformations, the equation will get into its simplified form then we can do the remaining calculation very easily.

Complete step by step answer:
Now, from the question, we have been given that \[{{5}^{x}}+1=625\]
Now, as we have already discussed above, we have to use some simple transformations to solve the given equation from the question.
Here, what we have to do is shift \[1\] from the left-hand side of the equation to the right-hand side of the equation.
By doing this, we get,
\[{{5}^{x}}+1=625\]
\[\Rightarrow {{5}^{x}}=625-1\]
\[\Rightarrow {{5}^{x}}=624\]
Now, what we have to do is, apply the logarithm with base \[5\] on both sides of the equation.
By applying the logarithm with base \[5\]on both sides of the equation, we get
\[{{\log }_{5}}{{5}^{x}}={{\log }_{5}}624\]
\[\Rightarrow x={{\log }_{5}}624\]
To find\[x\] we can try changing the base and use a pocket calculator that can give us the value.
Now, by changing base, we get
\[x={{\log }_{5}}624\]
\[\Rightarrow x=\dfrac{{{\log }_{10}}624}{{{\log }_{10}}5}\]
\[\Rightarrow x=3.999\]
So, as we have already discussed above, we got the solution by using some simple transformations.
Hence, the given question is solved.

Note:
We should be very careful while doing the calculation of the given question. We should be well aware of the transformations and substitutions that have to be made for the given question to get the equation more simplified. Once the equation gets simplified, we should do the calculation very carefully to get the given question solved. We should first understand the question correctly and then have to decide which transformation is to be used to get the question solved. The calculations present in the process of evaluation are as follows \[x=\dfrac{{{\log }_{10}}624}{{{\log }_{10}}5}\Rightarrow x=\dfrac{{{\log }_{10}}\left( 625-1 \right)}{{{\log }_{10}}5}\Rightarrow \dfrac{{{\log }_{10}}\left( {{5}^{4}}-1 \right)}{{{\log }_{10}}5}\] so the value of $ x $ will be greater than $ 3 $ and less than $ 4 $ .