Can the following system be solved algebraically?

2ln(x) + y= 4

e^x - y = 0

where ln(x) is loge(x)

If so, how would you do it?

Yes. You need to work out x in terms of y or y in terms of x in one equation, then substitute in the other equation. Or simply add the equations together: 2ln(x)+eˣ=4.

Draw a graph of 2ln(x)+eˣ-4. The graph crosses the x axis at around x=1.26, so this an approximate solution. [But without a graph, by substituting x=1 in the equation this function is negative, and by substituting x=2 it’s positive, so the solution lies between x=1 and 2. Continuing in this way the solution can be found (the solution lies between x=1.2 and 1.3 is the next step).]

Newton’s method can then be used to get a more accurate solution. Then y can be found from y=eˣ:

x[n+1]=x[n]−(2ln(x[n])+e^(x[n])-4)/(2/x[n]+e^(x[n])) where x[0]=1.26 and square brackets mean subscript. After only two or three iterations very accurate results are obtained, using a calculator, of course. Other algebraic solutions are possible using iterative procedures, independent of Newton’s method, but they may take longer to produce the same degree of accuracy.

Applying this method, x=1.262415239263 and y=3.53394651461 approx.

Another method is to solve for y by putting x=ln(y) in the first equation:

2ln(ln(y))+y=4, so y=4-2ln(ln(y)). If we start with y=e, then when we work out the right-hand side we get y=4. We then substitute y=4 in the right-hand expression and, using a calculator, we get y=3.3467... Now we put this result into the right-hand side again and the next value of y=3.6220... And so on. Eventually we get the value found above. To find x we use x=ln(y). This method is probably the easiest but it takes longer than Newton’s method.

by Top Rated User (764k points)