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27) ** solve: (left( \frac{1}{125} ight)^{2x - 3} = 25^{4x - 1})

Question

  1. ** solve: (left( \frac{1}{125}

ight)^{2x - 3} = 25^{4x - 1})

Explanation:

Step1: Express bases as powers of 5

Note that \( \frac{1}{125} = 5^{-3} \) and \( 25 = 5^{2} \). So rewrite the equation:
\( (5^{-3})^{2x - 3} = (5^{2})^{4x - 1} \)

Step2: Apply exponent power rule \((a^m)^n=a^{mn}\)

Simplify both sides:
\( 5^{-3(2x - 3)} = 5^{2(4x - 1)} \)
\( 5^{-6x + 9} = 5^{8x - 2} \)

Step3: Set exponents equal (since bases are equal)

If \( a^m = a^n \), then \( m = n \) (for \( a>0,a
eq1 \)). So:
\( -6x + 9 = 8x - 2 \)

Step4: Solve for x

Add \( 6x \) to both sides:
\( 9 = 14x - 2 \)
Add 2 to both sides:
\( 11 = 14x \)
Divide both sides by 14:
\( x = \frac{11}{14} \)

Answer:

\( x = \frac{11}{14} \)