Question #3f2bd

1 Answer
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