[RESOLVED] Exponential...Is this correct?

[Yunie]

Solve the equation:

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

Ans is y = 2 and y = -1

Is the answer correct? If not, could you please show me the correct workings for this question? Thanks.

[krtxmrtz]

Solve the equation:

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

Ans is y = 2 and y = -1

Is the answer correct? If not, could you please show me the correct workings for this question? Thanks.

I assume you've gone in the correct direction but you must work till the end and give x.

As in a previous question of yours, make a substitution:

e^2x = y

Then,

y - 2y^-1 = 1
or
y² - y -2 = 0

Solve for y to obtain: y = 2 and y = -1

Now, e^2x = y so x = (1/2) ln(y)

For y = -1, the ln is undefined. For y = 2,

x = (1/2) ln(2) which is the correct solution.

[Yunie]

Oops, I forgot. Haha. :P Thanks for correcting!