Solving an Exponential Equation with Base 2
To solve the equation \(4^{3x} = 32^{x-5}\), we can express both sides of the equation with a base of 2, as both 4 and 32 are powers of 2:
\(4 = 2^2\) and \(32 = 2^5\).
So, we can rewrite the equation as
\((2^2)^{3x} = (2^5)^{x-5}\).
Using the exponentiation rule \( (a^b)^c = a^{bc} \), we get
\(2^{6x} = 2^{5x - 25}\).
Since the bases are the same, we can set the exponents equal to each other:
\(6x = 5x - 25\).
Now, we solve for \(x\):
\(6x - 5x = -25\)
\(x = -25\).
Thus, the solution to the equation is \(x = -25\).
