Let's start by rewriting the equation so it's easier to read:
Prove that:
# (1 + log_5 8 + log_5 2) / (log_5 6400) = 0.5#
First, we know that #log_x a + log_x b = log_x ab#. We use that to simplify our equation:
# (1 + log_5 8 + log_5 2) / (log_5 6400) = (1 + log_5 (8*2)) / (log_5 6400) = (1 + log_5 16) / (log_5 6400)#
That "#1+#" is getting in the way, so let's get rid of it. We know that #log_x x = 1#, so we substitute:
#(1 + log_5 16) / (log_5 6400) = (log_5 5 + log_5 16) / (log_5 6400) #
Using the same addition rule from before, we get:
#(log_5 5 + log_5 16) / (log_5 6400) = (log_5 5 * 16) / (log_5 6400) = (log_5 80) / (log_5 6400)#
Finally, we know that #log_x a = log_b a / log_b x#. This is commonly called the "change of base formula" - an easy to way to remember where the #x# and #a# go is that #x# is below the #a# in the original equation (because it's written smaller under #log#).
We use this rule to simplify our equation:
#(log_5 80) / (log_5 6400) = log_6400 80#
We can re-write the logarithm into an exponent to make it easier:
#log_6400 80 = x#
#6400^x = 80#
And now we see that #x = 0.5#, since #sqrt(6400) = 6400^0.5 = 80#.
#square#
You probably made the mistake that #(log_5 80) / (log_5 6400) = 80/6400 = 1/80#. Be careful, this is not true.