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

1 Answer

Answer:

#x=log255~~2.407#

Explanation:

#-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#