How do you solve # 2^x = 20#?

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Answer 1 · 2016-02-29T23:16:08

Use basic properties of logarithms to remove #x# from the exponent to find that

#x = log_2(20)~~4.322#

We will use the following properties of logarithms:

With these, we have

#2^x = 20#

#=> log_2(2^x) = log_2(20)#

#=>xlog_2(2)=log_2(20)#

#=>x(1)=log_2(20)#

#:.x = log_2(20)~~4.322#