Answer :
Answer:
Proof below
Step-by-step explanation:
Irrational and Rational Numbers
We need to use the basic property of logarithms:
[tex]log_b\ a=x\ \ \Rightarrow b^x=a[/tex]
We are given:
[tex]log_3\ 7[/tex]
And we will prove it's an irrational number, i.e., it's not a rational number.
Recall a rational number can always be expressed as a fraction a/b, where a and b are integer numbers. Thus, if the expression was a rational number, then:
[tex]\displaystyle log_3\ 7=\frac{a}{b}[/tex]
Applying the property:
[tex]\displaystyle 3^{\frac{a}{b}}=7[/tex]
Raising to the power of b:
[tex]3^a=7^b[/tex]
It's not possible to find two integers a and b that satisfy the condition above. That contradiction comes from the assumption that [tex]log_3\ 7[/tex] is rational, thus, it's proven [tex]log_3\ 7[/tex] is irrational.