Question

Lg^5 x - 3lg^3 x - 4 lgx = 0

Answer 1

$lg^5x-3\, lg^3x-4\, lgx=0\; ,\; \; \; ODZ:\; x>0\\\\lgx\, (lg^4x-3\, lg^2x-4)=0\\\\a)\; \; lgx=0\; ,\; \; \underline {x=1}\\\\b)\; \; lg^4x-3lg^2x-4=0\\\\t=lg^2x\geq 0\; \to \; \; t^2-3t-4=0\; \; ,\; \; t_1=-1\; ,\; t_2=4>0,\\\\ t_1=-1$

Answer 2

To solve the equation lg^5(x) - 3lg^3(x) - 4lg(x) = 0, we can use a substitution to simplify it. Let's make a substitution:

Let y = lg(x).

Now, the equation becomes:

y^5 - 3y^3 - 4y = 0

This is a polynomial equation in terms of y. We can try to factor it or solve it by factoring out common terms:

y(y^4 - 3y^2 - 4) = 0

Now, we have two factors:

1. y = 0
2. y^4 - 3y^2 - 4 = 0

Let's solve each part separately:

1. y = 0 Since y = lg(x), we have: lg(x) = 0

   To solve for x, we need to remove the logarithm by exponentiating both sides: x = 10^0 x = 1

2. Now, let's solve the quartic equation: y^4 - 3y^2 - 4 = 0

   To solve this quartic equation, we can make a substitution. Let z = y^2:

   z^2 - 3z - 4 = 0

   Now, we have a quadratic equation in terms of z, which is easier to solve. We can factor it:

   (z - 4)(z + 1) = 0

   Now, solve for z:

   z = 4 or z = -1

   Since z = y^2, we have two possibilities:

   For z = 4: y^2 = 4

   Take the square root of both sides (considering both positive and negative roots): y = ±2

   For z = -1: y^2 = -1

   This equation has no real solutions because the square of a real number is always non-negative.

So, the solutions for y (and consequently x) are:

1. y = 2
2. y = -2
3. x = 10^2 = 100
4. x = 10^(-2) = 0.01

Therefore, the equation lg^5(x) - 3lg^3(x) - 4lg(x) = 0 has two real solutions: x = 100 and x = 0.01.