Question 258161
The original equation was
(i) {{{e^(2x)-2e^(x)-35=0}}}
Let Y = e^x. then Y^2 = e^(2x). We get
(ii) {{{Y^2-2Y-35=0}}}
factoring, we get
(iii) {{{(Y-7)(Y+5) = 0}}}
solving
Y = 7
OR
Y = -5
substituting e^x for Y, we get
(iv) {{{e^x = 7}}}
so, x = ln(7)
--
(v) {{{e^x = -5}}}
gives no solution. so, we are left with
x = ln(7)