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How do you solve ${5^x} = 30?$

Answer
VerifiedVerified
546k+ views
Hint:
Here they have asked us to find an exponent term. Whenever they ask us to find the exponent term of any number we have to make use of a logarithmic function that is by taking log on both the sides and try to simplify the obtained equation to get the correct answer.

Complete step by step solution:
From the given question that is ${5^x} = 30$ we can observe that the number $5$ is called the base and $x$ is called as the exponent.
So now we have to find the value of exponent that is $x$ .
In order to solve for exponent of any number we can make use of logarithmic function, that is by taking log on both side of the given equation ${5^x} = 30$
Therefore, on taking the log on both the sides, we get
$\log {5^x} = \log 30$
We can observe in the above equation that we have $\log {5^x}$ which is of the form $\log {a^b}$ . Whenever we have $\log {a^b}$ form we can write it as $b.\log a$ .
Hence we can write the above equation as,
$ \Rightarrow x\log 5 = \log 30$
Now, to make simplification easier, divide the both the sides of above equation by $\log 5$
$ \Rightarrow x\dfrac{{\log 5}}{{\log 5}} = \dfrac{{\log 30}}{{\log 5}}$
$ \Rightarrow x = \dfrac{{\log 30}}{{\log 5}}$
Now calculate the value of $\log 30$ and $\log 5$ by using a calculator.
So we have
$ \Rightarrow x = \dfrac{{1.47712}}{{0.69897}}$
$ \Rightarrow x = 2.11328$ which can be approximated and written as $x \approx 2.11$

Therefore the value of exponent in the given equation is $2.11$.

Note:
Whenever you are asking for an exponent try to take logs and simplify the expression. If you are not allowed to use a calculator to calculate logarithmic function then you try to multiply number two several times till you get the right hand side answer, but sometimes we do not get the exact number in this case. So it’s better to use a logarithmic function to solve for exponent. If you get the exponent value in terms of decimal as we got in this problem then try to approximate and write it rather than writing so many numbers after the decimal point.