Question

Lg^2(x) + lg(x^2) - lg^2(2) + 1 = 0

Answer 1

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Answer 2

lg(x)=y

y^2+2y+1=lg^2(2)

y+1=lg(2)  

или

y+1=-lg(2)

y=lg(0,2)

или

y=lg(0,05)

x=0,2  или  x=0,05   это ответ.

Answer 3

To solve the equation $\lg^2(x) + \lg(x^2) - \lg^2(2) + 1 = 0$, we need to work with the logarithms. Here, $\lg$ denotes the logarithm base 10. Let's simplify the equation step by step:

1. Simplify $\lg(x^2)$: Since $\lg(x^2) = 2 \cdot \lg(x)$, the equation becomes: $\lg^2(x) + 2 \cdot \lg(x) - \lg^2(2) + 1 = 0$

2. Let $y = \lg(x)$: The equation now becomes $y^2 + 2y - \lg^2(2) + 1 = 0$.

3. Substitute $\lg^2(2)$ with $y = \lg(2)$: Recall that $\lg(2) \approx 0.301$. So, $\lg^2(2) \approx (0.301)^2 \approx 0.0906$. The equation now becomes $y^2 + 2y - 0.0906 + 1 = 0$.

4. Now, solve for $y$ using the quadratic formula: $y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (1 - 0.0906)}}{2}$ $y = \frac{-2 \pm \sqrt{4 - 3.6384}}{2}$ $y = \frac{-2 \pm \sqrt{0.3616}}{2}$ $y = \frac{-2 \pm 0.6013}{2}$

5. Calculate the two possible values of $y$: a) $y = \frac{-2 + 0.6013}{2}$ which gives $y = -0.6993$ b) $y = \frac{-2 - 0.6013}{2}$ which gives $y = -1.3007$

6. Recall that $y = \lg(x)$, so we need to solve for $x$ in each case: a) $\lg(x) = -0.6993$ $x = 10^{-0.6993} \approx 0.2008$

   b) $\lg(x) = -1.3007$ $x = 10^{-1.3007} \approx 0.0506$

So, the solutions for the equation are approximately $x = 0.2008$ and $x = 0.0506$.