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3 years ago

6

How do you solve the function y = 2.5^x for y=20

1 answer:

Andre45 [30]3 years ago

6 0

$y=2.5^x;\ y=20\\\\2.5^x=20\\\\log_{2.5}2.5^x=log_{2.5}20\\\\x=log_{2.5}20\\\\x=log_{2.5}(2.5\cdot8)\\\\x=log_{2.5}2.5}+log_{2.5}8\\\\x=1+log_{2.5}8$

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Answer:

1) $x < 0$

2) $x > 0$

3) $x < 10$

4) $x < -\frac{31}{55}$

5) $x \geq 0$

6) $x \geq -1$

Step-by-step explanation:

1) $-3 > 2x -3$

$2x- 3 < -3$

$2x - 3 + 3 < -3 +3$

$2x < 0$

$\frac{2x}{2} < \frac{0}{2}$

$x < 0$

2) $5 > -x + 5$

$-x + 5 < 5$

$- x + 5 -5 < 5 - 5$

$-x < 0$

$\frac{-x}{-1} < \frac{0}{-1}$

$x > 0$

3) $\frac{x}{5} -2 < 0$

$5 \times (\frac{x}{5} - 2 ) \times 5 < 0$

$x - 10 < 0$

$x -10 + 10 < 0 +10$

$x < 10$

4) $-5x + 11 > 31$

$-5x \cdot 11 < 31$

$-55x < 31$

$\frac{-55x}{-55} > \frac{31}{-55}$

$x < -\frac{31}{55}$

5) $\frac{1}{2}x - 4 \geq -4$

$\frac{1}{2}x -4 +4 \geq -4 +4$

$\frac{1}{2}x \geq 0$

$\frac{\frac{1}{2}x }{\frac{1}{2} } \geq \frac{0}{\frac{1}{2} }$

$x \geq 0$

6) $-4x + 2 \leq 6$

$-4x +2 - 2 \leq 6 -2$

$-4x < 4$

$\frac{-4x}{-4} \leq \frac{4}{-4}$

$x \geq -1$

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3 years ago