An exponential equation such as [tex]$5^x = 9[tex]$[/tex] can be solved for its exact solution using the meaning of logarithms and the change-of-base theorem. Because [tex]$[/tex]x[tex]$[/tex] is the exponent to which 5 must be raised in order to obtain 9, the exact solution is [tex]$[/tex]\log_5 9[/tex], or [tex]$\frac{\log 9}{\log 5}[tex]$[/tex], or [tex]$[/tex]\frac{\ln 9}{\ln 5}$[/tex].

For the following equation, give the exact solution in three forms similar to the forms above:
[tex]\[6^x = 30\][/tex]

For the given equation, the exact solutions in three forms are:
[tex]\[ \log_6 30, \frac{\log 30}{\log 6}, \frac{\ln 30}{\ln 6} \][/tex]



Answer :

Let's solve the exponential equation \(6^x = 30\) and express the exact solution in three different logarithmic forms, similar to how \(\log_5 9\) can be expressed as \(\frac{\log 9}{\log 5}\) or \(\frac{\ln 9}{\ln 5}\).

1. Exact log form:
The logarithmic form of the solution involves using the base 6 logarithm, representing the power to which 6 must be raised to get 30:
[tex]\[ x = \log_6 30 \][/tex]

2. Change-of-base formula (using common logarithm \(\log\)):
We can apply the change-of-base theorem, which allows us to convert the logarithm into any base. The common logarithm (base 10) is often used:
[tex]\[ x = \frac{\log 30}{\log 6} \][/tex]

3. Change-of-base formula (using natural logarithm \(\ln\)):
We can also use the natural logarithm (base \(e\)) to express the solution:
[tex]\[ x = \frac{\ln 30}{\ln 6} \][/tex]

Therefore, the exact solutions for the equation \(6^x = 30\) in the three different forms are:

1. \(\log_6 30\)
2. \(\frac{\log 30}{\log 6}\)
3. \(\frac{\ln 30}{\ln 6}\)

Hence, filling the exact solution forms into the provided form looks like:

For the given equation, the exact solutions in three forms are [tex]\(\log_6 30\)[/tex], [tex]\(\frac{\log 30}{\log 6}\)[/tex], and [tex]\(\frac{\ln 30}{\ln 6}\)[/tex].