Question

4^x + 6^x = 9^x. Find the value of x ??

Answer 1

x = log( (1  + √5)/2 ) / log(3/2)   or  x  = log ((-1 + √5)/2)/log(2/3)

Step-by-step explanation:

4ˣ  + 6ˣ   = 9ˣ

Dividing both sides by 4ˣ

=> 1  +  (6/4)ˣ  = (9/4)ˣ

=> 1  + (3/2)ˣ  = ((3/2)²)ˣ

=>  ((3/2)²)ˣ -  (3/2)ˣ - 1 = 0

=> ((3/2)ˣ )² -  (3/2)ˣ - 1 = 0

Let say (3/2)ˣ = y

=> y² - y - 1 = 0

=> y  =  (1  ± √5)/2

(3/2)ˣ  = (1  ± √5)/2

Taking log both sides

x log(3/2)  = log ((1  + √5)/2 )    as log of -v2 not defined

=> x = log( (1  ± √5)/2 ) / log(3/2)

or

4ˣ  + 6ˣ   = 9ˣ

Dividing both sides by 9ˣ

=> (4/9)ˣ  +  (6/9)ˣ  = 1

=> ((2/3)ˣ)²  + (2/3)ˣ - 1 = 0

=> (2/3)ˣ   =  (-1 ± √5)/2

=> x  = log ((-1 + √5)/2)/log(2/3)

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