How to solve an issue with an equation that has a variable with multiple values.

I have this equation

𝑦 = 𝑒^(3𝑥)(2𝑥+𝑙𝑛(40𝑥))

and have to solve y using the values x:[1,2,3,4,5,6,7,8,9,10]

I currently have this as my code, but I keep getting wrong values for y. I am just wondering where I went wrong.

x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

y = (exp(3*x).*((2*x)+(log(40*x))))

here are my results

y =

1.0e+14 *

0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0003 0.0058 0.1271 2.7776

Answer 1

[Dyuman Joshi]

You are not getting wrong values of y.

format long
x = (1:10)';
y = (exp(3*x).*((2*x)+(log(40*x))))

y = 10×1

1.0e+14 * 0.000000000001143 0.000000000033816 0.000000000874120 0.000000021280472 0.000000500104652 0.000011477782122 0.000258946694719 0.005766237134192 0.127085596247646 2.777571252192637

Since there is a big difference in the first value and the last value of the expression, and when presented in short format (the default format), the answer is displayed with (first) 4 (rounded) digits in exponent/scientific notation

When some of the values in an array are short numbers and some have large exponents (as is your case), you can use long format as shown above or shortG as shown below

format shortG
y

y = 10×1

1.0e+00 * 114.26 3381.6 87412 2.128e+06 5.001e+07 1.1478e+09 2.5895e+10 5.7662e+11 1.2709e+13 2.7776e+14

[Javier Solis]

Thank you, I greatly appreciate it.