e^(ln 3) = 3
e1 + (lnx) = e1 * e(lnx) = e * x = ex
0.8333
Think of ln 1 as "e to what power will given me 1." Anything to the zero power will give you 1. So, ln 1 = 0, and 0/2 = 0
integral of e to the power -x is -e to the power -x
1/10 to the 7th power equals 1,000,000
e^(ln 3) = 3
10-7 = 0.0000001
If you're multiplying, it's a to the sixteenth. If you're dividing, it's a to the second.
e1 + (lnx) = e1 * e(lnx) = e * x = ex
5/x4
0.8333
There are 2 interpretations of your question: First: e^[lnx + lny] =e^[ln(xy)] =xy Second: lny + e^(lnx) =lny + x
e2xlnx = x2x since ln x is the opposite to e so you just leave the x from the log and the rest of the exponent stays the same.
Cube root is the same as to the power of a third; when multiplying/dividing powers of a number add/subtract the powers; when a power is to another power, multiply the powers; as it is all e to some power: e³/(e²)^(1/3) × e^13 = e³/e^(2/3) × e^13 = e^(3 - 2/3 + 13) = e^(15 1/3) = e^(46/3) Which can also be expressed as "the cube root of (e to the power 46)" or "(the cube root of e) to the power 46".
You can't simplify that expression, unless you know specific values for the variables.
e times 5 = X