ok, apparently seeing the e^2x is making this problem seem too complicated to you. let's replace e^2x with y and see if you follow.

let e^2x be y, then we have

y^2 - y - 30 = 0 .............does it look simplier now? it should, this is a regular (and simple) quadratic equation.

=> (y - 6)(y + 5) = 0

=> y = 6 or y = -5

but y is e^2x

=> e^2x = 6 or e^2x = -5

e^2x can never be -5 so there is no solution for that equaition, so

=> e^2x = 6 ............this is an exponential equaiton, we proceed by logging both sides

=> lne^2x = ln6

=> 2x = ln6

=> x = (ln6)/2

this is log to the base 10 i assume

2. log(4x) - log(12+ square root of x) = 2

to me that just looked flat out intimidating =/

log(4x) - log(12 + sqrt(x)) = 2 .........let's combine the logs on the left

=> log[(4x)/(12 + sqrt(x))] = 2 ........since logx - logy = log(x/y)

=> 10^2 = (4x)/(12 + sqrt(x)) .........since if log[a] b = c, then a^c = b (log[a]b means log to the base a of b)

=> 100 = 4x/(12 + sqrt(x))

=> 120 + 100sqrt(x) = 4x

=> 4x - 100sqrt(x) - 120 = 0 .........that itself is complicated! Now what?!!

calm down, let's rewrite this

4x - 50*2sqrt(x) - 120 = 0 ..........are you seeing it yet? it's similar to the first one

=> (2sqrt(x))^2 - 50(2sqrt(x)) - 120 = 0 .........see it now?

let 2sqrt(x) be y

=> y^2 - 50y - 120 = 0

for some reason i'm having trouble seeing the factors for this, let's use the quadratic formula

=> y = [50 +/- sqrt(2500 + 480)]/2

=> y = [50 +/- sqrt(2980)]/2 ...........no wonder i could not see the factors, there's no nice number for that!

but y = 2sqrt(x)

=> 2sqrt(x) = [50 +/- sqrt(2980)]/2

but 2sqrt(x) can't be negative, so we only want the positive answer, that is,

2sqrt(x) = [50 + sqrt(2980)]/2

=> sqrt(x) = 50 + sqrt(2980)

=> x = [50 + sqrt(2980)]^2

so just work that out