Question #3f2bd

1 Answer
smartspot2
Jan 6, 2018

#x < log_10(4)#
or
#x < 0.60206#

Explanation:

First, subtracting 5 from all sides:

#4 < 10^x + 5 < 9#

#4-5 < 10^x + 5 - 5 < 9-5#

#-1 < 10^x < 4#

We can take the #log_10# of all the sides to bring down the #x#:

#log_10(-1) < x < log_10(4)#

However, #log_10(-1)# is undefined; #log_10(x)# only exists for #x ge 0#. So, the original question is kind of misleading; #10^x+5# only gives values for #x ge 5#, so it would be unnecessary to put 4 on the left hand side.

Since #log_10(x)# goes towards #-\infty# as #x\rightarrow0#, we can just replace #log_10(-1)# with #-infty#:

#-infty < x < log_10(4)#

The #-infty# is now redundant, because that doesn't place any lower bound on #x#, so we can remove that entirely, giving us our answer:

#x < log_10(4)#

#x < 0.60206#

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