logarithm question, PLEASE HELP, i'd appreciate it!?

ex - e-x = 2

ex - 1/ex = 2

lcm

e2x - 1 = 2ex

e2x - 2ex -1 =0

put ex =t

t^2 -2t - 1 = 0

t = {-(-2) +- √(-2^2 - 4(1)(-1))}/2(1)

t = {2 +- √ (4+4)}/2

t = 1+- √2

so

ex= 1 + √2 and ex = 1-√2

so x = ln (1+√2)

and

x = ln (1-√2) (neglect this as ln argument cant be negative )

so answer is x = ln(1+√2)

x = ln(1+1.414)

= ln(2.414)

= 0.88137

(assets :-- log a N =z so N=a^z) log9 27=x 27=9^x (making use of the above assets) 3^3=3^2x by using fact the backside are comparable i.e=3 so equating the powers 3=2x x=3/2 log2 8=y 8=2^y (lower back making use of the valuables) 2^3=2^y lower back base is comparable i.e 2 so equating the powers y=3 log9 27 . log2 8=x*y=3/2*3=9/2

ex = 2 + e + x

x = (2 + e + x)/e

Maybe you meant: e^x - e^(-x) = 2.

e^(2x) - 1 = 2e^x

e^2x - 2e^x - 1 = 0

Let u = e^x

u^2 - 2u - 1 = 0

u^2 - 2u = 1

u^2 - 2u + 1 = 1 + 1

(e^u - 1)^2 = 2

e^u - 1 = ± √2

e^u = 1 ± √2

u = ln (1 + √2), because u - √2 < 0 and isn't in the domain of ln.

u = 1 + √2 = e^x

x = ln (1 + √2)

If it is e^x-e^(-x)=2

then let e^x=y

y^2-2Y-1=0

y=1+2^(1/2)

Therefore x=log(1+sqrt(2))

Do you mean e^x - e^(-x) = 2?