How do you solve for x in #9=2^(.84x)#?

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Answer 1 · 2015-12-05T15:38:25

#x~~3.77# to 2 decimal places

Suppose we had #log(x^3)#

This is #log(x xx x xx x )#

Which is #log(x)+log(x)+log(x)#

Which is #3log(x)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given: #9=2^(0.84x)#

Take logs of both sides

#log(9) = 0.84xlog(2)#

#x=(log(9))/(0.84xxlog(2))#

#x~~3.77# to 2 decimal places