What are all roots of the equation (0.6^x)*(25/9)^(x^2-12)=(27/125)^3

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve: (0.6^x)*(25/9)^(x^2-12)=(27/125)^3

(0.6^x)*(25/9)^(x^2-12)=(27/125)^3

=> (3/5)^x*(5/3)^2^(x^2 - 12) = (3^3/5^3)^3

=> (3/5)^x*(5/3)^2^(x^2 - 12) = (3/5)^9

=> (3/5)^x*(3/5)^2^(12 - x^2) = (3/5)^9

=> (3/5)^(x + 2*(12 - x^2)) = (3/5)^9

As the base is the same equate x + 24 - 2x^2 = 9

x + 24 - 2x^2 = 9

=> 2x^2 - x -15 = 0

=> 2x^2 - 6x + 5x - 15 = 0

=> 2x(x - 3) + 5(x - 3) = 0

=> (2x + 5)(x - 3) = 0

=> x = -5/2 and x = 3

The solution of the equation is x= -5/2 and x = 3

 

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the 1st factor as a fraction:

0.6^x = (6/10)^x = (3/5)^x

We notice that 25/9 = (5/3)^2

We'll raise both sides by (x^2 - 12):

(25/9)^(x^2 - 12)  = (5/3)^2*(x^2 - 12)

We notice that 27/125 = (3/5)^3

We'll raise both sides by 3:

(27/125)^3 = (3/5)^9

We'll write the equation:

[(3/5)^x]*[(5/3)^2*(x^2 - 12)] = (3/5)^9

But [(5/3)^2*(x^2 - 12)] = [(3/5)^-2*(x^2 - 12)]

Since the bases form the left side are matching, we'll add the exponents:

[(3/5)^(x-2*(x^2 - 12))] = (3/5)^9

Since the bases form the left side are matching, we'll apply one to one rule:

x-2*(x^2 - 12) = 9

We'll remove the brackets:

x - 2x^2 + 24 - 9 = 0

-2x^2 + x + 15 = 0

We'll calculate the roots of the equation:

x1 = [-1+sqrt(1+120)]/-4

x1 = (-1+11)/-4

x1 = -5/2

x2 = 3

The required roots of the equation are: {-5/2 ; 3}.

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