Solving the limit as x approaches infinity for y = 10 * e ^ ((ln (1/2)/4)*x)?

Question

Is it zero?

and how do you get it?

Answer 1

ln (1/2) = ln1 - ln2 = -ln2

So we're looking at e^((a negative number)*x/4)

Doesn't matter the number-it's negative and doesn't approach zero. Infinity times a negative number and divided by four is negative infinity, so we're looking at 10*e^(-inf). That's the same as 10/e^inf, which does approach zero.

Answer 2

If you have 10* e^ln[(1/2)/4*x] by definition of logarithm
e^lna=a.In this case you have lim 10*(1/8)*x which tends to infinity with the sign of x

If you have 10* e^[ln(1/8) *x] you can write it( operating rules with exponents) as10*[e^ln(1/8)]^x which is 10*(1/8)^x .
So if x =>+ infinity the limit is 0 because 1/8<1
If x=>-infinity 1/8^x = 8^-x and -x tends to +infinity and 8^-x also tends to + infinity.( the factor 10 has no influence)