# What is the solution of the equation (3/5)^6(x+1)-(27/125)^(x-3)=0?

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### 2 Answers

The equation we have to solve is : (3/5)^6(x+1)-(27/125)^(x-3)=0

(3/5)^6(x+1)-(27/125)^(x-3)=0

=> (3/5)^6(x+1)-(3/5)^3^(x-3)=0

=> (3/5)^6(x+1)-(3/5)^(3x - 9)=0

=> (3/5)^6(x+1) = (3/5)^(3x - 9)

As the base is the same equate the exponent

6(x + 1) = 3x - 9

=> 6x + 6 = 3x - 9

=> 3x = -15

=> x = -5

**The solution of the equation (3/5)^6(x+1)-(27/125)^(x-3)=0 is x = -5**

For the given expression to become an equation, you'll have to write:

(3/5)^6(x+1)-(27/125)^(x-3) = 0

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

Writing 27/125 as (3/5)^3, we'll create matching bases.

The equation will become:

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

We'll shift (3/5)^3(x-3) to the right:

(3/5)^6(x+1) = (3/5)^3(x-3)

Since the bases are matching, we'll apply one to one rule of exponentials:

6(x+1) = 3(x-3)

We'll divide by 3:

2(x+1) = x - 3

We'll remove the brackets:

2x + 2 = x - 3

We'll isolate x to the left side:

2x - x = -2 - 3

x = -5

**The solution of the equation is x = -5.**