Find greatest value of $y(x) = (0.9^x)(300x + 650)$ Question and attempt 

$y(x) = (0.9^x)(300x + 650)$

Estimate at what x value that y reaches its maximum value

The only way I could think of would be to use derivatives, so I tried it:
$y'(x) = (0.9^x)' \times (300x + 650) + 0.9^x \times (300x + 650)'$
          $=$ [$\ln(0.9) \times 0.9^x \times 1$] $\times  (300x + 650) + 0.9^x \times 300$
So then I subbed in 0 to find where the turning point is:
$y(0) =\ln0.9 \times 0.9^0 \times (300(0) + 650) + 0.9^0 \times 300$
        $=\ln0.9 \times 650 + 300$
        ~ $231.51$
The problem is that $231.51$ is not the correct answer.

The real answer
This is a table of values of the answer (see that it increases up to in between x = 7 and 8, then decreases):
x = 5: 1269.5535
x = 6: 1302.03045
x = 7: 1315.316475
x = 8: 1312.9249905
x = 9: 1297.85863815

Here is what the graph actually looks like (which tells me there is a turning point):

And here is the picture of the vertex:

So the x value at which the y value is at its maximum value is $7.3$
I'd like to know where I went wrong with the derivative idea, but I still would encourage answers describing any other way to solve the problem.

You were asked to give an approximation, not an exact answer. If you take the derivative and compare it to zero, you get the rather difficult expression:
$$(300x+650)[\ln(0.9)×0.9^x]+0.9^x×300=0$$
We know $\ln (1+x) \sim x$ for small $x$. So, we get:
$$(300-30x-65)0.9^x=0$$
This will only be zero when $235=30x$, or when $x$ is approximately $7.8$
