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garyljc
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I was wondering if anyone could show me why b^[log (base b) a ] is = a ?
D H said:The equation
[tex]b^{\log_b a} = a[/tex]
is better written as
[tex]b^{\log_b a} \equiv a[/tex]
In other words, the equality is true by definition.
garyljc said:is it possible to prove it ? like simplify it then showed that it is equal to a ?
The expression b^[log (base b) a ] is equivalent to a because the logarithm base b of a is the power to which b must be raised to equal a. In other words, the logarithm base b "undoes" the exponentiation of b, leaving us with just a as the final result.
The base of the logarithm does not affect the value of b^[log (base b) a ]. This is because the logarithm base b is the inverse function of exponentiation with base b. In simpler terms, the base of the logarithm cancels out the base of the exponentiation, leaving us with only a as the final result.
Let's say we have the expression 3^[log (base 3) 9 ]. The logarithm base 3 of 9 is 2, because 3^2 = 9. Therefore, the expression simplifies to 3^2, which is equal to 9. In this example, the logarithm base 3 "undoes" the exponentiation of 3, leaving us with the original value of 9.
Yes, b^[log (base b) a ] is always equal to a for any positive values of b and a. This is because the logarithm and exponentiation are inverse operations of each other, meaning they cancel each other out and leave us with the original value of a.
The concept of b^[log (base b) a ] = a is a fundamental property of logarithms and exponentials. It illustrates how the logarithm and exponentiation are inverse operations of each other, and how the base of the logarithm and exponentiation cancel each other out. This property is also known as the "logarithm power rule" and is essential in solving equations and manipulating expressions involving logarithms and exponentials.