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]$[/tex]\log_5 9[tex]$[/tex], or [tex]$[/tex]\frac{\log 9}{\log 5}[tex]$[/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]8^x = 26[/tex]

For the given equation, the exact solutions in three forms are:

1. [tex]\log_8 26[/tex]
2. [tex]\frac{\log 26}{\log 8}[/tex]
3. [tex]\frac{\ln 26}{\ln 8}[/tex]

(Type integers or fractions.)



Answer :

To solve the equation \( 8^x = 26 \) for \( x \), we will use logarithms and the change-of-base theorem. Let's go through the solution step by step.

1. Logarithmic Form:

The solution \( x \) is the power to which 8 must be raised to get 26. In logarithmic form, we express this as:
[tex]\[ x = \log_8 26 \][/tex]
So the first form of the solution is:
[tex]\[ \log_8 26 \][/tex]

2. Using the Common Logarithm (base 10) and the Change-of-Base Theorem:

The change-of-base theorem states that for any positive numbers \( a \), \( b \), and \( c \) (with \( a \neq 1 \) and \( b \neq 1 \)):
[tex]\[ \log_b a = \frac{\log_c a}{\log_c b} \][/tex]
Using this theorem with common logarithms (base 10), we get:
[tex]\[ x = \frac{\log 26}{\log 8} \][/tex]
So the second form of the solution is:
[tex]\[ \frac{\log 26}{\log 8} \][/tex]

3. Using the Natural Logarithm (base \( e \)):

Similarly, we can use the natural logarithm (base \( e \)) as well:
[tex]\[ x = \frac{\ln 26}{\ln 8} \][/tex]
So the third form of the solution is:
[tex]\[ \frac{\ln 26}{\ln 8} \][/tex]

Therefore, the exact solutions for \( x \) in the equation \( 8^x = 26 \) can be written in three forms:

1. \(\log_8 26\)
2. \(\frac{\log 26}{\log 8}\)
3. \(\frac{\ln 26}{\ln 8}\)

These represent the three equivalent ways to express the solution.