Question

Solve for x: 4^x+6^x=9^x

Answer 1

Answer:

this is the correct answer

Answer 2

QUESTION -

Solve for x: 4^x+6^x=9^x

ANSWER -

4^x + 6^x - 9^x = 0

Divide both sides by 9^x:

(4/9)^x + (6/9)^x -1 = 0

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

Substitute t = (2/3)^x, t > 0.

t^2 + t - 1 = 0

Let's solve equation:

D = (-1)^2 - 4*1*(-1) = 1 + 4 = 5

t = (-1 +/- √5)/2

t1 = (-1 - √5)/2 is negative, so doesn't fit

t2 = (-1 +√5)/2 > 0

Now let's get back to substitution:

t = (2/3)^x = (-1 +√5)/2

Take natural logarithm of both sides

ln[ (2/3)^x] = ln[(-1 +√5)/2]

Applying properties of logarithms:

x*ln(2/3) = ln[(√5 - 1)/2]

x = ln[(√5 - 1)/2] / ln(2/3) <== The answer