3^(2x−1)= 1/(27^x)
5^(3x−8)·5^(4x)=25^(2x)
this one also stumped me
27 = 3^3, so
3^(2x−1)= 1/(27^x)
3^(2x−1)= 1/3^(3x)
3^(2x-1) = 3^(-3x)
2x-1 = -3x
5x = 1
x = 1/5
25=5^2, so
5^(3x−8)·5^(4x) = 25^(2x)
5^(3x−8)·5^(4x) = 5^(4x)
3x-8+4x = 4x
x = 8/3
thanks you Steve!
:) :)
To solve the equation 3^(2x−1) = 1/(27^x), we need to simplify both sides of the equation and then find the value of x that satisfies the equation.
Let's start by simplifying the equation:
First, we can rewrite 1/(27^x) as 3^(-3x), since 27 is equal to 3^3.
So, the equation becomes: 3^(2x−1) = 3^(-3x)
Now, we can equate the exponents:
2x−1 = -3x
Next, we'll solve for x. To do this, we'll isolate the terms with x on one side of the equation:
2x + 3x = 1
Combining the x terms, we get:
5x = 1
Finally, we solve for x by dividing both sides of the equation by 5:
x = 1/5
So, the solution to the equation 3^(2x−1) = 1/(27^x) is x = 1/5.