Question

(lg(x)^2)^2+lg10x-6=0

Answer 1

$(lgx^2)^2+lg10x-6=0$
ОДЗ:  $x\ \textgreater \ 0$
$(2lgx)^2+(lg10+lgx)-6=0$
$4lg^2x+1+lgx-6=0$
$4lg^2x+lgx-5=0$
Замена:  $lgx=a$
$4a^2+a-5=0$
$D=1^2-4*4*(-5)=81$
$a_1= \frac{-1+9}{8} =1$
$a_2= \frac{-1-9}{8} =-1.25$
$lgx=1$       или        $lgx=-1.25$
$x=10$        или        $x=10^{- \frac{5}{4} }= \frac{1}{10^{ \frac{5}{4}} }= \frac{1}{ \sqrt[4]{10^5} } =\frac{1}{ 10\sqrt[4]{10} }$

Ответ:  $\frac{1}{ 10\sqrt[4]{10} } ;$  $10$

Answer 2

The equation you've provided is a quadratic equation with respect to the variable lg(x) (the base-10 logarithm of x). Let's solve it step by step:

Given equation: (lg(x)^2)^2 + lg(10x) - 6 = 0

Let's simplify the equation: Let y = lg(x)^2 So, the equation becomes: y^2 + lg(10x) - 6 = 0

Now we have a quadratic equation in terms of y: y^2 + lg(10x) - 6 = 0

To solve for y, we can use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / 2a

In this case, the coefficients are: a = 1 (coefficient of y^2) b = lg(10x) c = -6

Plug these values into the quadratic formula: y = (-lg(10x) ± √((lg(10x))^2 - 4 * 1 * -6)) / (2 * 1) y = (-lg(10x) ± √(lg(10x)^2 + 24)) / 2

Now remember that we defined y = lg(x)^2. So, substitute y back: lg(x)^2 = (-lg(10x) ± √(lg(10x)^2 + 24)) / 2

Now, solve for lg(x): lg(x) = ±√((-lg(10x) ± √(lg(10x)^2 + 24)) / 2)

Remember that the logarithm of a negative number is not defined in the real number system, so the solutions will be valid only for values of x that make the expression inside the square root non-negative.

Please note that solving this equation analytically might not lead to simple closed-form solutions due to the presence of logarithms. You might need to use numerical methods or calculators to find approximate solutions for specific values of x.