Question

How do you solve `5^(x+3)>10^(x-6)`?

Answer 1

`x < 6+9log_2 5`

Explanation:

`10^(x-6)=2^(x-6)5^(x-6)` then

`5^(x+3)/(5^(x-6)) > 2^(x-6)` but

`5^(x+3)/(5^(x-6)) =5^(x+3-(x-6)) = 5^9` then

`5^9 > 2^(x-6)` or

`5^9 2^6 > 2^x` now aplying `log` to both sides the inequality is preserved because `log` is monotonic strict increasing. So

`9log_2 5+6log_2 2>x` or finally

`x < 6+9log_2 5`