# Solve the equations 3^(3x-3)*5^(x-4)=15^2x/5^7

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to solve 3^(3x-3)*5^(x-4)=15^2x/5^7

3^(3x-3)*5^(x-4)=15^2x/5^7

=> (3^3x / 3^3)*(5^x / 5^4) = (5*3)^2x / 5^7

=> (3^3x / 27)*(5^x) = (5*3)^2x / 5^3

=> (3^3x / 27)*(5^x) = (5*3)^2x / 125

=> (3^3x / 3^2x)(5^x / 5^2x) = 27/125

=> 3^x / 5^x = (3/5)^3

=> (3/5)^x = (3/5)^3

As the base is the same we can equate the exponent

=> x = 3

The required value of x is x = 3

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that we can write 15 = 3*5 => 15^2x = 3^2x*5^2x

We'lldivide 15^2x by 5^7:

15^2x/5^7 = 3^2x*5^2x/5^7

We'll use qutotient property for the exponentials that have matching bases:

3^2x*5^2x/5^7 = 3^2x*5^(2x-7)

We'll re-write the equation:

3^(3x-3)*5^(x-4)=3^2x*5^(2x-7)

We'll divide by 3^2x and we'll get:

3^(3x-3)*5^(x-4)/3^2x = 5^(2x-7)

We'll divide by 5^(x-4):

3^(3x-3)/3^2x = 5^(2x-7)/5^(x-4)

We'll subtract the exponents:

3^(3x - 3 - 2x) = 5^(2x - 7 - x + 4)

We'll combine like terms inside brackets:

3^(x - 3) = 5^(x - 3)

We'll re-write the equation:

3^x*3^-3 = 5^x*5^-3

3^x/3^3 = 5^x/5^3

We'll create matching bases. We'll divide by 5^x:

3^x/5^x*3^3 = 1/5^3

We'll multiply by 3^3:

3^x/5^x = 3^3/5^3

(3/5)^x = (3/5)^3

Since the bases are matching, we'll apply one to one property:

x = 3

The real solution of the equation is x = 3.