Question

solve this real quick​

Answer 1

Let

2^x = a

and

3^x = b

So,

9^x = (3^2)^x = (3^x)^2 = b^2

6^x = (3 · 2)^x = 3^x · 2^x = ab

4^x = (2^2)^x = (2^x)^2 = a^2

Now,

9^x + 6^x = 2 · 4^x

=> b^2 + ab = 2a^2

=> 2a^2 - ab - b^2 = 0

=> (2a^2 - ab - b^2)/b^2 = 0/b^2

=> 2a^2/b^2 - ab/b^2 - b^2/b^2 = 0

=> 2(a/b)^2 - a/b - 1 = 0

Taking a/b = k,

=> 2k^2 - k - 1 = 0

=> 2k^2 - 2k + k - 1 = 0

=> 2k(k - 1) + (k - 1) = 0

=> (k - 1)(2k + 1) = 0

=> k = 1 ; k = -1/2

=> a/b = 1 ; a/b = -1/2

=> (2^x)/(3^x) = 1 ; (2^x)/(3^x) = -1/2

=> (2/3)^x = 1 ; (2/3)^x = -1/2

From (2/3)^x = 1, we get,

x = log 1 to the base (2/3)

x = 0

[Because log 1 to the base a = 0, since a^0 = 1.]

(2/3)^x = -1/2 is impossible because the LHS is positive since 2/3 is positive but the RHS is negative.

Hence x has only a unique value and that is none other than 0.