Maximum of 2^x - 3^x?

Question

How can I find the maximum of this? Taking the derivative results in 2^x * ln2 - 3^x *ln3, which I don't know how to solve for 0.

Answer 1

Well, you've already got the derivative, so set it equal to zero:

2^x ln 2 - 3^x ln 3 = 0

Add 3^x ln 3 to both sides:

2^x ln 2 = 3^x ln 3

Now, divide both sides by 2^x ln 3:

ln 2/ln 3 = (3/2)^x

Now, simply take the log of both sides:

ln (ln 2/ln 3) = x ln (3/2)

Divide by ln 3/2:

ln (ln 2/ln 3)/ln (3/2) = x

Evaluating this mess numerically yields:

x ≈ - 1.135882567916304

Answer 2

you could graph it

2^x ln2 =3^xln3
ln(2^(2^x))=ln(3^(3^x))
2^(2^x)=3^(3^x)
( i dont know where im going with this. It looks like you would have to graph it).

Answer 3

there is no maximum because there are no critical points. there are no values for x that make it equal to 0 or undefined (any number to the power of x can never equal 0, and there is no denominator so it cannot be undefined), so the only possible "maximum" is an endpoint. you can always graph it too

Answer 4

2^x (ln 2) - 3^x (ln 3) = 0

2^x (ln 2) = 3^x (ln 3)

log [2^x (ln 2)] = log [3^x (ln 3)]

log 2^x + log(ln 2) = log3^x + log(ln 3)

x log 2 + log(ln 2) = x log 3 + log(ln 3)

x log 2 - x log 3 = log(ln 3) - log(ln 2)

x(log 2 - log 3) = log(ln 3) - log(ln 2)

x = [log(ln 3) - log(ln 2)]/(log 2 - log 3)

Good luck plugging that back into the original equation though. :c)

Answer 5

f(x)^x = e^(xln(f(x)))

2^x = e^(xln(2)) and 3^x = e^(xln(3))

y = 2^x + 3^x
dy/dx = ln(2)2^x - ln(3)3^x

up to this point, you've done everything right!

when dy/dx = 0, (take ln() of both sides)
ln(2)2^x = ln(3)3^x
ln(ln(2)) + ln(2^x) = ln(ln(3)) + ln(3^x)
ln(ln(2)) + xln(2) = ln(ln(3)) + xln(3)

x = [ln(ln(3)) - ln(ln(2))]/[ln(2) - ln(3)]
= ln[ln(3)/ln(2)]/ln(2/3)
= ln[ln(3-2)]/ln(2/3)
= 0 no surprise!