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