Dividing both sides by xthroot(16). It'll become xth root(36/16) + xth root(24/16) = 1, xth root(9/4) + xth root(3/2) = 1. {xth root(3/2)}^2 + xth root(3/2) = 1. Put xth root(3/2) as t. So equation becomes t^2 + t - 1 = 0. Then solve for t and from there calculate x.
Dividing both sides by xthroot(16). It'll become xth root(36/16) + xth root(24/16) = 1, xth root(9/4) + xth root(3/2) = 1. {xth root(3/2)}^2 + xth root(3/2) = 1. Put xth root(3/2) as t. So equation becomes t^2 + t - 1 = 0. Then solve for t and from there calculate x.
I am glad that I have reviewed this problem. And I shall use that for practice!!!
(9/4)^x + (6/4)^x =1
(9/4)^x + (3/2)^x = 1
Let y =( 3/2 )^x
y^2 + y - 1 =0
y = 1/2(-1+/-(1+4)^1/2
(3/2)^x = 1/2(-1+/-5^1/2)
x =( 2/3)* log (-1+/-5^1/2)/2
(36/16)^(1/x)+(24/16)^(1/x)=((3/2)^2)^(1/x)+(3/2)^(1/x) , ((3/2)^2)^(1/x)+(3/2)^(1/x)=1 , let u=(3/2)^(1/x) , u^2+u-1=0 ,
u= (-1+V5)/2 , /(-1-V5)/2 < 0 , not a solu / , (3/2)^(1/x)=(-1+V5)/2 , x=ln(3/2)/ln((-1+V5)/2) , x=~ -0.842592 ,
test , 36^(1/x)+24^(1/x)=~ 0.37233 , 16^(1/x)=~ 0.37233 , same , OK ,