equation raising a number to x

Status
Not open for further replies.

headman333

Member level 5
Joined
Jan 19, 2008
Messages
82
Helped
1
Reputation
2
Reaction score
2
Trophy points
1,288
Activity points
1,769
3^x + 9^x = 27^x , Anybody knows how to solve for x?
 

I have substituted that in the equation but i find no way to find x.
 

Use the hint given by SunnySkyguy. At this point you'll have:

3^x+3^(2x)=3^(3x)

that is:

3^x+(3^x)^2=(3^x)^3

we can write y=3^x, then:

y+y^2=y^3 much simpler, now.

rearranging:

y*(1+y-y^2)=0

this means we have three solutions:

y1=0
y2=[1+sqrt(5)]/2
y3=[1-sqrt(5)]/2

we wrote y=3^x, so inverting the function x=log3 (that is logarithm base 3 of y)

Then y1 and y3 are not valid since they aren't strictly positive. The solution will be:

x=log3{[1+sqrt(5)]/2} = 0.438....
 


Thanks! I finally got the right answer.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…