Solving Logarithms: Discovering the Unknown Variable in Log_a (100)

[Luis Melo]

How do i solve Log_a (100) if Log_a (2) = 20 and Log_a (5) = 30

I got to 2^(1/20) = 100^(1/x) and 5^(1/30) = 100^(1/x) but didnt know how to go any further.

 

[Ray Vickson]

How do i solve Log_a (100) if Log_a (2) = 20 and Log_a (5) = 30

I got to 2^(1/20) = 100^(1/x) and 5^(1/30) = 100^(1/x) but didnt know how to go any further.

If log_a(x) means log to base "a" of x, then the two conditions you gave are inconsistent: you get two different values of "a" in the two cases.

 

[pasmith]

How do i solve Log_a (100) if Log_a (2) = 20 and Log_a (5) = 30

I got to 2^(1/20) = 100^(1/x) and 5^(1/30) = 100^(1/x) but didnt know how to go any further.

Is your question "What is $\log_a(100)$ if $\log_a(2) = 20$ and $\log_a(5) = 30$"?

Well, you can get the answer from the fundamental principle of logarithms: $\log_a(xy) = \log_a(x) + \log_a(y)$.

However this is a spectacularly poorly designed question, since it asserts that $a^{20} = 2$ and $a^{30} = 5$, which together require $5^2 = 2^3$. This is plainly false.

 

[Luis Melo]

Thank you or the answers.