How would you solve for x? For some reason I just can't figure them out... I greatly appreciate anyone who could help me :)

#1) y= √x

y = x^2

#2) y=e^-x

y=e^x

#3) y=4-x^2

y=1-x^2

#4)

y=e^x

y=ex

#5) y=x^3-3x+1

x+y

Printable View

- Sep 12th 2009, 05:28 PMloutja35How to solve for x?
How would you solve for x? For some reason I just can't figure them out... I greatly appreciate anyone who could help me :)

#1) y= √x

y = x^2

#2) y=e^-x

y=e^x

#3) y=4-x^2

y=1-x^2

#4)

y=e^x

y=ex

#5) y=x^3-3x+1

x+y - Sep 12th 2009, 06:15 PMmathaddict
Hi

(1) $\displaystyle y^2=x$ ---1 (square both sides)

$\displaystyle y=x^2 $---2

sub 2 into 1

$\displaystyle (x^2)^2=x $

$\displaystyle

x^4=x

$

$\displaystyle x^4-x=0$ .... continue here

(2) $\displaystyle y^x=e$ --1

$\displaystyle y=e^x$ ---2

sub 2 into 1

$\displaystyle (e^x)^x=e $

$\displaystyle e^{2x}=e $

continue here

Can you try the rest ? - Sep 13th 2009, 06:39 AMloutja35
Although you said continue here and I can see your process....I don't know how to continue forward. It looks to me like x isn't being solved for. I don't understand what I am doing wrong. Please help...thank you to anyone who can

- Sep 13th 2009, 07:23 AMstapel
1) Factor. Apply

**the difference-of-cubes formula**to x^3 - 1. Solve the two linear factors.

2) To learn how to solve exponential equations, try**here**. (Wink)