Let's solve each equation one by one.
Equation 1:
\[-x^{-2} = -\frac{1}{81}\]
Multiply both sides by -1: \[x^{-2} = \frac{1}{81}\]
This implies: \[\frac{1}{x^2} = \frac{1}{81}\]
Taking the reciprocal of both sides: \[x^2 = 81\]
Taking the square root of both sides: \[x = 9 \quad \text{or} \quad x = -9\]
Equation 2:
\[x^{\frac{3}{4}} = 27\]
To solve for \(x\), raise both sides to the power of \(\frac{4}{3}\): \[(x^{\frac{3}{4}})^{\frac{4}{3}} = 27^{\frac{4}{3}}\]
This simplifies to: \[x = 27^{\frac{4}{3}}\]
Calculating \(27^{\frac{4}{3}}\): First, find \(27^{\frac{1}{3}} = 3\). Then raise it to the power of 4: \[3^4 = 81\]
Thus, \[x = 81\]
Equation 3:
\[25 \cdot 5^{x-1} = 24\]
Divide both sides by 25: \[5^{x-1} = \frac{24}{25}\]
Taking the logarithm (base 5) of both sides: \[x - 1 = \log_5\left(\frac{24}{25}\right)\]
Hence, \[x = \log_5\left(\frac{24}{25}\right) + 1\]
To simplify further, Using the change of base formula: \[x = \frac{\log\left(\frac{24}{25}\right)}{\log(5)} + 1\]
Now we've solved all the equations. The final values are:
- \(x = 9\) or \(x = -9\)
- \(x = 81\)
- \(x = \log_5\left(\frac{24}{25}\right) + 1\)
These are the solutions for the respective equations.