How do you solve #-16+0.2(10^x)=35#?

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Answer 1 · 2016-10-07T12:39:31

#x=log255~~2.407#

#-16+0.2(10^x)=35#

Let's move all the non-x terms to the right side:

#0.2(10^x)=51#

#10^x=51/0.2=255#

Now we can take the log of both sides:

#log10^x=log255#

#xlog10=log255#

Remember that #Log_(10)10=1#, so:

#x=log255~~2.407#