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Vlad [161]
3 years ago
15

3.3^(2x+1)-103^x+1=0 need value of x

Mathematics
1 answer:
photoshop1234 [79]3 years ago
4 0

Answer:

The value of x is approximately -1.531.

Step-by-step explanation:

Let 3.3^{2\cdot x + 1}-103^{x+1} = 0, we proceed to solve this expression by algebraic means:

1) 3.3^{2\cdot x + 1}-103^{x+1} = 0  Given

2) 3.3^{2\cdot x}\cdot 3.3 -103^{x}\cdot 103 = 0 a^{b}\cdot a^{c} = a^{b+c}

3) (3.3^{x})^{2}\cdot 3.3 -\left[\left( \sqrt{103} \right)^{2}\right]^{x}\cdot 103 = 0 (a^{b})^{c} = a^{b\cdot c}

4) (3.3^{x})^{2}\cdot 3.3 - \left[\left(\sqrt{103}\right)^{x}\right]^{2}\cdot 103 = 0 (a^{b})^{c} = a^{b\cdot c}/Commutative property

5) \left[\left(\frac{3.3}{\sqrt{103}}\right)^{x}\right] ^{2}-\frac{103}{3.3} = 0 Existence of multiplicative inverse/Definition of division/Modulative property/a^{b}\cdot a^{c} = a^{b+c}

6) \left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\frac{103}{3.3} Existence of additive inverse/Modulative property/(a^{b})^{c} = a^{b\cdot c}

7) \log \left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\log \frac{103}{3.3} Definition of logarithm.

8) 2\cdot x\cdot \log \left(\frac{3.3}{\sqrt{103}} \right)= \log \frac{103}{3.3}     \log_{b} a^{c} = c\cdot \log_{b} a

9) 2\cdot x \cdot [\log 3.3-\log \sqrt{103}] = \log 103 - \log 3.3      \log_{b} \frac{a}{d}

10) x\cdot (2\cdot \log 3.3-\log 103) = \log 103 - \log 3.3     \log_{b} a^{c} = c\cdot \log_{b} a/Associative property

11) x = \frac{\log 103-\log 3.3}{2\cdot \log 3.3-\log 103}   Existence of multiplicative inverse/Definition of division/Modulative property

12) x \approx -1.531  Result

The value of x is approximately -1.531.

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