Question

4^x+9^x=6^x then x=?​

Answer 1

Step-by-step explanation:

Hello,

Here, 4^x+9^x=6^x

=> 1 + (9/4)^x =(6/4)^x

=> 1+ (3/2)^2x= (3/2)^x

=>(3/2)^2x -(3/2)^x +1=0..... (1)

putting t=(3/2)^x in eqn (1)

then, t^2=(3/2)^2x

=> t^2 -t+1=0

=> t = [-(-1)+(1-4)^1/2]/2

or, [-(-1)-(1-4)^1/2]/2

so,(3/2)^x= t = [1+i(3)^1/2]/2 or, [1-i(3)^1/2]

Now, taking logarithm both sides we get:-

x log(3/2)=log [{1+i(3)^1/2}/2]

Hence,

x = log [{1+i (3)^1/2}/2]/log(3/2) ans..

Thank you! !