Prove that #(1 + Log_5 8 + Log_5 2)/log_5 6400 = 0.5# Please note the base number of each log is 5 and not 10. I continuously get #1/80#, can someone please assist?

2 Answers
Feb 23, 2018

#1/2#

Explanation:

#6400 = 25*256 = 5^2*2^8#
#=> log(6400) = log(5^2) + log(2^8) = 2 + 8 log(2)#
#log(8) = log(2^3) = 3 log(2)#
#=> (1+log(8)+log(2))/log(6400) = (1+4 log(2))/(2+8log(2)) = 1/2#

Feb 23, 2018

Apply common logarithmic identities.

Explanation:

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.