Question

What is the largest integer `x`, for which the value of `f(x)=5x^4+30x^2+9` will be greater than the value of `g(x)=3^x`?

Answer 1

`x=9`

Explanation:

We're looking for the largest integer where:

`f(x)>g(x)`

`5x^4+30x^2+9>3^x`

There are a few ways we can do this. One is to simply try out integers. As a baseline, let's try `x=0`:

`5(0)^4+30(0)^2+9>3^0`

`0+0+9>1`

and so we know that `x` is at least 0 so there's no need to test negative integers.

We can see that the largest power on the left is 4. Let's try `x=4` and see what happens:

`5(4)^4+30(4)^2+9>3^4`

`5(256)+30(4)^2+9>81`

I'll hold off on the rest of the math - it's clear the left side is bigger by a considerable amount. So let's try `x=10`

`5(10)^4+30(10)^2+9>3^10`

`5(10000)+30(100)+9>59049`

`50000+3000+9>59049`

so `x=10` is too big. I think our answer will be 9. Let's check:

`5(6561)+30(81)+9>19683`

`32805+30(81)+9>19683`

and again it's clear the left side is bigger than the right. So our final answer is `x=9`.

What are other ways to find this? We could have tried graphing. If we express this as `(5x^4+30x^2+9)-3^x=0`, we get a graph that looks like this:

graph{(5x^4+30x^2+9)-3^x [0, 11, -10000, 20000]}

and we can see that the answer peaks around the `x=8.5` mark, is still positive at `x=9` and turns negative before reaching `x=10` - making `x=9` the biggest integer.

How else could we do this? We could solve `(5x^4+30x^2+9)-3^x>0` algebraically.

`5x^4+30x^2+9-3^x>0`

To make the math easier, I'm first going to notice that as the values of `x` increase, the left side terms start to become irrelevant. First the 9 will decrease in significance until it's completely irrelevant, and the same goes for the `30x^2` term. So this reduces to:

`5x^4>3^x`

`log(5x^4)>log(3^x)`

`4log5x>xlog3`

`4log5+4logx>xlog3`

`(4log5+4logx)/log3>x`

and I think I'm making a mess of this! algebra is not an easy way to approach this problem!