Question

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

Answer 1

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.