Solve for x the equation 2^2x=6^(x+1) - 5*3^2x?  

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll move all terms to the left side:

2^2x - 6^(x+1) + 5*3^2x = 0

We'll write 6^x = 2^x*3^x, also 6^(x+1) = (6^x)*6

6^(x+1) = 6*2^x*3^x

We'll re-write the equation:

2^2x - 6*2^x*3^x + 5*3^2x = 0

We'll divide by 3^2x:

(2/3)^2x - 6*(2/3)^x + 5 = 0

We'll substitute (2/3)^x = t

t^2 - 6t + 5 = 0

We'll apply quadratic formula:

t1 = [6 + sqrt(36-20)]/2

t1 = (6+4)/2

t1 = 5

t2 = (6-4)/2

t2 = 1

We'll substitute (2/3)^x = 1, where 1 = (2/3)^0

(2/3)^x = (2/3)^0 <=> x = 0

(2/3)^x = 5 <=> x*ln (2/3) = ln 5

x = ln 5/ln(2/3)

The values of x that represent the solutions of the equation are: x = 0 and x = (ln 5)/ln(2/3).

We’ve answered 318,955 questions. We can answer yours, too.

Ask a question