solve for x : (3/2)^x - (2/3)^x = 1?

p = (3/2)^x

p - 1/p = 1

p² - p - 1 = 0

p = (1 +√5) / 2

(3/2)^x = (1 +√5) / 2

x ln (3/2) = ln ((1 +√5)/2)

Answer:

x = ln((1 + √5)/2) / ln(3/2) =1.1868143902809817175449880

When all else fails, try a little t&e (trial and error)

Actually at x=1, you're not too far away from an answer, so try somthing around x=4/3. You can also do this with logs (remember that the log(2/3) is a negative number. )

x = [lg (1 + √5) - 1] / [lg 3 - 1] ≈ 1.187

(lg means log base 2)

Let A = (3/2)^x, so

A - A⁻¹ = 1

A² - 1 = A

A² - A - 1 = 0

A = (1 ± √[(-1)² - 4(1)(-1)])/2(1)

A = ½(1 ± √5)

(3/2)^x = ½(1 ± √5)

x lg [3/2] = lg [½(1 ± √5)]

x = lg[½(1 ± √5)] / lg[3/2]

x = [lg (1 ± √5) - lg 2] / [lg 3 - ln 2]

x = [lg (1 ± √5) - 1] / [lg 3 - 1]

We can discount the minus if the answer is to be real, since 1 - √5 is negative, and the lg of a negative number is undefined over the real numbers.

Let y = (3/2)^x... then

y - 1/y = 1

→ y = phi or y=-1/phi where phi is the Golden Ratio

Solution 1: y = (3/2)^x = phi

→ x = log phi / (log 3 - log 2)

Solution 2: y = (3/2)^x = -1/phi doesn't work because

LHS > 0 > RHS.