# What is x if 7^(3x-5) = 1/49?

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We have to solve for x given that 7^(3x-5) = 1/49.

Now 7^(3x-5) = 1/49

=> 7^(3x-5) = 1/ 7^2

=> 7^(3x-5) = 7^-2

Now the base 7 is the same for terms on both the sides, we can equate the exponential.

=> 3x - 5 = -2

=> 3x = -2 + 5

=> 3x = 3

=> x = 3/3

=> x = 1

**Therefore the required value of x is 1.**

To find x if 7^(3x-5) = 1/49.

We notice that that the left side is expressed in terms of exponents of the base 7 and the right side is not expressed like that. So we write the right side 1/49 = 1/7^2 = 7^-2.

So given equation is now rewritten as:

7^(3x-5) = 7^(-2). So the bases being equal , exponents are also equal:

Therefore 3x-5 = -2. We solve this linear equation now.

3x-5 = -2.

We add 5 to both sides:

3x-5+5 = -2+5.

3x = 3.

3x/3 = 3/3.

x = 1.

So x= 1 is the solution.

We notice that the bases of exponentials from both sides could be written as powers of 7.

1/49 = 1/7^2

We'll use the negative power rule:

1/7^2 = 7^-2

We'll re-write the equation having common bases both sides:

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

Since the bases are matching, we'll use the one to one property of exponentials and we'll get:

(3x-5) = -2

We'll remove the brackets from the left side and add 5 both sides:

3x = 5 - 2

We'll combine like terms:

3x = 3

We'll divide by 3:

**x = 1**

**The valid solution of the equation is x = 1.**