My Name is Earl said:I have tried various methods to solve this...
If logb(a) = loga(b) where a != b (!= means does not equal), ab > 0 and neither a nor b are 1, then what is the value of ab?
MarkFL said:Let's let:
\(\displaystyle x=\log_a(b)=\log_b(a)\)
Now this implies:
\(\displaystyle a^x=b\)
\(\displaystyle b^x=a\)
Dividing the former by the latter, we obtain:
\(\displaystyle \left(\frac{a}{b}\right)^x=\left(\frac{a}{b}\right)^{-1}\)
What does this imply?
My Name is Earl said:This implies that ab = 1
The equation log_b(a) = log_a(b) means that the logarithm of a with base b is equal to the logarithm of b with base a. In other words, the two expressions are equivalent.
To solve for ab, you can use the property of logarithms which states that log_b(a) = x is equivalent to b^x = a. In this case, b = a and a = b, so you can rewrite the equation as a^x = b and solve for x. The solution will be ab = a * b = a^x * b^x = (ab)^x, so ab = 1.
Yes, the equation log_b(a) = log_a(b) is always true. This is because the two expressions are inverses of each other, meaning that if you apply one to a number, you can reverse the process by applying the other to get back to the original number. In this case, applying log_b(a) and then log_a(b) will bring you back to a, and vice versa.
The equation log_b(a) = log_a(b) has several applications in mathematics and science. It can be used to solve exponential and logarithmic equations, as well as to convert between different bases. It also has practical uses in fields such as finance and computer science.
Yes, the equation log_b(a) = log_a(b) can be extended to other bases. In general, the equation log_b(a) = log_c(a) / log_c(b) holds true for any base c. This is known as the change of base formula and is frequently used in solving logarithmic equations.