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Solve the inequality: $27^x-4\cdot125^x>45^x$

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Writing from expanded to exponent form

Full solution

Q. Solve the inequality:

27^x-4\cdot125^x>45^x

Rewrite bases as powers: Rewrite the bases as powers of a common base: $\newline$ $27 = 3^3$ , $125 = 5^3$ , $45 = 9 \times 5 = 3^2 \times 5$ . $\newline$ So, $27^x = (3^3)^x = 3^{(3x)}$ , $125^x = (5^3)^x = 5^{(3x)}$ , and $45^x = (3^2 \times 5)^x = 3^{(2x)} \times 5^x$ .
Substitute into inequality: Substitute these into the inequality: $3^{3x} - 4 \times 5^{3x} > 3^{2x} \times 5^x$ .
Divide and simplify: Divide all terms by $5^x$ to simplify: $\newline$ $\left(\frac{3^{3x}}{5^x}\right) - 4 \times \left(\frac{5^{3x}}{5^x}\right) > \frac{3^{2x} \times 5^x}{5^x}$ , $\newline$ which simplifies to $\frac{3^{3x}}{5^x} - 4 \times 5^{2x} > 3^{2x}.$
Rewrite with single exponents: Rewrite the terms with single exponents: $\newline$ $(3^3)^x / 5^x - 4 \times (5^2)^x > (3^2)^x$ , $\newline$ which simplifies to $3^{3x - x} - 4 \times 25^x > 9^x$ .

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