SOLUTION: Let E(x) = 30(1.05)^x and L(x) = 30 + 2x. I need to give an example of a value x when L(x) < E(x). Can someone please help? I am really struggling with this one. I have the res

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Let E(x) = 30(1.05)^x and L(x) = 30 + 2x. I need to give an example of a value x when L(x) < E(x). Can someone please help? I am really struggling with this one. I have the res      Log On


   



Question 695850: Let E(x) = 30(1.05)^x and L(x) = 30 + 2x.

I need to give an example of a value x when L(x) < E(x). Can someone please help? I am really struggling with this one. I have the rest of the problem figured out.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
L%28x%29+=+30+%2B+2x is a cleverly named highlight%28L%29inear function, while
E%28x%29+=+30%281.05%29%5Ex is a cleverly named highlight%28E%29xponential function.

Linear functions graph as straight lines, and L%28x%29+=+30+%2B+2x represents a line with a slope of 2 throughout its all real numbers domain.

Exponential functions graph as curved lines, and if the base is greater than one, as for E%28x%29+=+30%281.05%29%5Ex, have "exponential growth" graphs that look like this graph%28200%2C200%2C-0.99%2C0.99%2C-0.99%2C0.99%2C0.2%2A2%5E%282x%29%29, with slopes that increase from very small to very large.

The line could be tangent to the exponential curve and the graphs could intersect at only one point, as in graph%28200%2C200%2C-0.99%2C0.99%2C-0.99%2C0.99%2C0.6x%2B0.18%2C0.2%2Ae%5E%282x%29%29
Otherwise, a linear curve and an exponential curve could intersect at two points graph%28200%2C200%2C-0.99%2C0.99%2C-0.99%2C0.99%2C0.3x%2B0.3%2C0.2%2A2%5E%282x%29%29, or zero points graph%28200%2C200%2C-0.99%2C0.99%2C-0.99%2C0.99%2C0.3x%2B0.1%2C0.2%2A2%5E%282x%29%29.

In any of those cases, the line is below the exponential curve (L%28x%29+%3C+E%28x%29) at most (if not all) points, for most (if not all) values of x.

Which of those scenarios is the case for your functions?
L%280%29=30%2B2%2A0=30%2B0=30 and
E%280%29=30%281.05%29%5E0=30%2A1=30, so the graphs intersect at a least one point, with x=0.
We can look at points to either side.
If the curves are tangent at x=0, for all other values of x
it will be (L%28x%29+%3C+E%28x%29.
If the curves are not tangent at x=0, then they intersect at two points,
In that case, L%28x%29%3CE%28x%29 will be true at least in some interval to one side of x=0, and we may be lucky enough to find a point there.

For x=1:
L%281%29=30%2B2%2A1=30%2B2=32 and
E%281%29=30%281.05%29%5E1=30%2A1.05=31.5 so for x=1, L%28x%29%3EE%28x%29

For x=-1:
L%28-1%29=30%2B2%2A%28-1%29=30-2=28 and
E%281%29=30%281.05%29%5E%28-1%29=30%2F1.05=200%2F7=approx.28.57 so for highlight%28x=-1%29, L%28x%29%3CE%28x%29