Thread: Values of 2^x

For what values of 2^x does

2^x+2^x+1=4^x+4^x+1

Originally Posted by Dragon

For what values of 2^x does

2^x+2^x+1=4^x+4^x+1

Let y=2^x

Thus,

y+y+1=y^2+y^2+1

You shall solve.

Originally Posted by ThePerfectHacker

Let y=2^x

Thus,

y+y+1=y^2+y^2+1

You shall solve.

i keep on getting a negative square root??

Originally Posted by Dragon

For what values of 2^x does

2^x+2^x+1=4^x+4^x+1

Using TPH's method:

Let y = 2^x

Then
y + y + 1 = y^2 + y^2 + 1

2y = 2y^2 <-- Cancel the common factor of 2

y^2 - y = 0

y(y - 1) = 0

So y = 0 or y = 1.

Thus 2^x = 0 <-- Impossible!

or

2^x = 1 <-- x = 0

So the only solution is x = 0.

-Dan

Thanks alot

Hello, Dragon!

Some parentheses and spaces would have helped.
I'm guessing what the problem is supposed to be . . .

For what values of 2^x does: .2^x + 2^(x+1) .= .4^x + 4^(x+1)

If I've interpreted the problem correctly, we get an entirely different answer.


We have: .2^x + 2·2^x .= .(2²)^x + (2²)^(x+1}

. . 3·2^x .= .2^(2x) + 2^(2x + 2)

. . 3·2^x .= .2^(2x) + 2²·2^(2x)

. . 3·2^x .= .2^{2x} + 4·2^(2x)

. . 3·2^x .= .5·2^(2x)

Divide by 2^x: .3 .= .5·2^x . **

And we have: .2^x .= .3/5

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** . We can divide by 2^x because 2^x ≠ 0.