How do you evaluate #log_15(15)#?

1 Answer
Dec 3, 2015

Answer:

#log_a(a) = 1# holds for any positive integer #a#.

Explanation:

You are basically searching for #x# so that

#log_(15)(15) = x#

The inverse function of the #log_15(z)# is #15^z#which means that
#log_15(15^z) = z# and also # 15^(log_15(z)) = z# always hold.

This means that to "get rid" of the logarithmic term, you can perform # 15^z# on both sides of your equation:

#color(white)(xxx)15^(log_15(15)) = 15^x#

#<=> color(white)(xxxxx) 15 = 15^x#

#<=> color(white)(xxxxx) 15^1 = 15^x#

So you can see that #x = 1# which means that

#log_15(15) = 1#