ATTN: Math geniuses

[Staiain]

Can anyone solve this and show how you did the equations?






My friend needs help with this.

[hi19hi19]

For the bottom one, I believe you use the identity

Also I can give a general tip for dealing with e: sometimes it's helpful to think of it just as any old number, like say 2.7
Even though e has some special properties especially when dealing with logarithms, treating it as "just a number" can sometimes remind you of ways you can shuffle it around to solve your equation.
(this can also be a helpful mindset for working with pi, phi, really any irrational number)

Wish I had more time to go in-depth but if nobody else has responded by tonight I'll try to help better

[iironiic]

For the first one try factoring e^(1/2*x). It should be straightforward from there. I provided a solution in case you managed to work it out.

Factor out an e^(1/2) so that you get e^(1/2 *x) (e^(1/2*x) -2) = 0. e^(1/2 *x) > 0 so we need to consider e^(1/2*x) - 2 = 0, or e^(1/2*x) = 2. Take the natural log, you get 1/2*x = ln 2 and therefore x = 2 ln 2.

Second one. I don't understand the question. Is it ln (3*x) or 3 ln x? What is ln x3?

[leonid]

x3

[Reincarnate]

but srsly:

e^x - 2e^(.5x) = 0
e^x = 2e^(.5x)
e^x/(e^(.5x)) = 2
e^(.5x) = 2
.5x = ln(2)
x = 2ln(2)

Assuming the second question's rhs is actually ln(3x)

4 - ln(x^3/3) = ln(3x)
4 = ln(3x) + ln(x^3/3)
4 = ln(3x * x^3/3)
4 = ln(x^4)
e^4 = x^4
x = e

[emerald000]

I guess the second one is ln(x)^3.

4 - ln(x^3/3) = ln(x^3)
4 - ln(x^3) + ln(3) = ln(x^3)
4 + ln(3) = 2 ln(x^3)
(4 + ln(3)) / 2 = ln(x^3)
e^((4 + ln(3)) / 2) = x^3
x = e^((4 + ln(3)) / 6)
x = e^(2/3 + ln(3)/6)
x = e^(2/3) * e^(ln(3)/6)
x= e^(2/3) * 3^(1/6)