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Lynna [10]
3 years ago
13

What is the value of n when (9)^2n-1 =(27)^n+2

Mathematics
2 answers:
Sedaia [141]3 years ago
5 0

Answer:

n = 8

Step-by-step explanation:

Using the rule of exponents

(a^m)^{n} = a^{mn}

Given

(9)^{2n-1} = (27)^{n+2}, then

(3^2)^{2n-1} = (3^3)^{n+2}

3^{4n-2} = 3^{3n+6}

Since the bases on both sides are equal, both 3 then equate the exponents

4n - 2 = 3n + 6 ( subtract 3n from both sides )

n - 2 = 6 ( add 2 to both sides )

n = 8

LenKa [72]3 years ago
3 0

Answer:

n = 8

Step-by-step explanation:

9²ⁿ⁻¹ = 27ⁿ⁺²

(3²)²ⁿ⁻¹ = (3³)ⁿ⁺²

3⁴ⁿ⁻² = 3³ⁿ⁺⁶

4n - 2 = 3n + 6

4n - 3n = 6 + 2

n = 8

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Solve for x write the smaller solution first and the larger solution second. Round to two decimal places (x+8)^2-7=0
siniylev [52]

Answer:

x ≈ - 10.65, x ≈ - 5.35

Step-by-step explanation:

Given

(x + 8)² - 7 = 0 ( add 7 to both sides )

(x + 8)² = 7 ( take the square root of both sides )

x + 8 = ± \sqrt{7} ( subtract 8 from both sides )

x = - 8 ± \sqrt{7} , thus

x = - 8 - \sqrt{7} ≈ - 10.65

x = - 8 + \sqrt{7} ≈ - 5.35

4 0
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STatiana [176]

Rationalize the numerator:

\dfrac{\sqrt{x+4}-2}x\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac x{x(\sqrt{x+4}+2)}=\dfrac1{\sqrt{x+4}+2}

This is continuous at x=0, so we can evaluate the limit directly by substitution:

\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}x=\lim_{x\to0}\frac1{\sqrt{x+4}+2}=\frac1{\sqrt4+2}=\frac14

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