2500(1.25)^x=6000(0.3)^x
You asked:
Solve the equation \(2500 \cdot {1.25}^{x} = 6000 \cdot {0.3}^{x}\) for the variable \(x\).
MathBot Answer:
The solution is: \[x = - \frac{- \log{\left(6000 \right)} + 2 \log{\left(50 \right)}}{- \log{\left(3 \right)} - 2 \log{\left(2 \right)} + \log{\left(5 \right)} + \log{\left(10 \right)}} \approx 0.61345295\]