equation raising a number to x

[headman333]

3^x + 9^x = 27^x , Anybody knows how to solve for x?

 

[D.A.(Tony)Stewart]

hint
9^x=3^2x
27^x=3^3x

 

[headman333]

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

 

[Dominik Przyborowski]

Use logarithm function

 

[albbg]

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....

 

[headman333]

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.