log_a(36) - log_a(4) = 0.5, what is a?
This is an example of a question you might face during a higher maths examination. It is also a great question to test what your understanding of logarithms and powers are. I have given a quick explanation below which could be extended if a student got stuck.
To do this questions we must know a few things. First, we can simplify the left hand side of this equation such that:
LHSloga(36) - loga(4) = loga(36/4) = loga(9)
Here we had to remember that when we subtract two logarithms, this is equivalent to taking the logarithm of the division of these numbers. Now that we have simplified the left hand side of the equation, we can now re-write the problem as:
loga(9) = 0.5
We must remember that a logarithm and a power are inverse operations of one another (simplified explanation). In the end we can get rid of a logarithm by taking it to the power of its base. As usual, any operation we do has to be done to both sides of the equation giving:
alog_a(9) = a0.5
9 = a0.5
This trick allows us to get rid of the logarithm and now simply calculate for a. In order to get rid of the power to one half which we have on the rhs, we can square both sides of the equation giving:
92 = (a0.5*a0.5)
which givesa = 81.
