How do you solve #2^x=3.7#?

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Answer 1 · 2015-10-20T18:15:05

#x=log3.7/(log2)# or #x~~1.89#

take the #log# of both sides:

#log2^x=log3.7#

There is a rule that states that #loga^b# can be written as #b*loga#

#xlog2=log3.7#

#x=log3.7/(log2)# (you can use your calculator to solve this)

#x~~1.89#

Note:
You can also use #ln# instead of #log#; your answer should be the same either way