Exponential Equation: 2^x^2-3 = 64

[jenjen]

Im having trouble with this problem. The answer is 3,-3
here is the problem

2^x^2-3=64

 

[stapel]

2^x^2-3=64

Your formatting is ambiguous. Do you mean any of the following?

. . . . .2[sup:18f2wpmm]x^(2-3)[/sup:18f2wpmm] = 64

. . . . .2[sup:18f2wpmm]x^2 - 3[/sup:18f2wpmm] = 64

. . . . .(2[sup:18f2wpmm]x[/sup:18f2wpmm])[sup:18f2wpmm]2[/sup:18f2wpmm] - 3 = 64

. . . . .2[sup:18f2wpmm]x^2[/sup:18f2wpmm] - 3 = 64

Or something else? What are your thoughts? What have you tried? How far have you gotten?

When you reply, kindly use the formatting explained in the links in the "Read Before Posting" post you read before posting. Thank you!

Eliz.

 

[jenjen]

Sorry I meant the following.
2x^2 - 3 = 64
the x^2-3 is all raised

this is how I far i get on the problem
2x^2 - 3 = 2^6
x^2-3=6
x^2=9

at this point should I use the difference of two squares?

 

[stapel]

Sorry I meant the following.
2x^2 - 3 = 64
the x^2-3 is all raised

So, if you were going to use the requested formatting, you would type the exercise as follows...?

. . . . .Solve 2^(x^2 - 3) = 64.

In other words: \(\displaystyle \mbox{Solve }\, 2^{x^2\, -\, 3}\, =\, 64.\)

this is how I far i get on the problem
2x^2 - 3 = 2^6
x^2-3=6
x^2=9
at this point should I use the difference of two squares?

You can do that:

. . . . .x[sup:1pj43imi]2[/sup:1pj43imi] = 9
. . . . .x[sup:1pj43imi]2[/sup:1pj43imi] - 9 = 0
. . . . .(x - 3)(x + 3) = 0

...and solve the factors. Or you could just go straight to taking the square roots:

. . . . .\(\displaystyle \sqrt{x^2}\, =\, x\, =\, \pm\sqrt{9}\)

...and so forth.

Eliz.

 

[jenjen]

Thanks for your help!