How do you solve #5^x= 20#?

1 Answer
Dec 6, 2015

Take logs and use properties of logs to find:

#x = (log 20)/(log 5) = (1+log 2)/(1 - log 2) ~~ 1.86135#

Explanation:

If we take common logs of both sides then we get:

#log 20 = log 5^x = x log 5#

So #x = (log 20)/(log 5)#

We can do a little more with this if we know the log value #log 2 ~~ 0.30103# and that #log 10 = 1#:

#log 5 = log (10/2) = log 10 - log 2 = 1 - log 2#

#~~ 1 - 0.30103 = 0.69897#

#log 20 = log (10*2) = log 10 + log 2 = 1 + log 2#

#~~ 1 + 0.30103 = 1.30103#

So #x = (log 20)/(log 5) ~~ 1.30103 / 0.69897 ~~ 1.86135#