Question

$\log_x (125x) \cdot \log_{25}^{2} x = 1$ $\log_x 5^3x \cdot \log_{5^2}^{2} x = 1$ $\log_x 5^3x = \frac{1}{\log_{5^2}^{2} x}$ $\log_x 5^3x = \log_{x}^{2} 5^2$ $5^3x =$

Answer 1

$x=5, \frac{1}{625}$

Answer 2

Solution:

The given equation is $\log_x (125x) \cdot \log_{25}^{2} x = 1$.

First, we determine the domain for which the logarithms are defined. For $\log_x (125x)$, we must have the base $x > 0$ and $x \neq 1$, and the argument $125x > 0$. Since $125 > 0$, $x > 0$. For $\log_{25} x$, we must have the base $25 > 0$ and $25 \neq 1$ (which are true), and the argument $x > 0$. Combining these conditions, the domain is $x > 0$ and $x \neq 1$.

Now, we simplify the terms using logarithm properties, preferably converting to a common base, such as base 5. $\log_x (125x) = \frac{\log_5 (125x)}{\log_5 x} = \frac{\log_5 (5^3 \cdot x)}{\log_5 x} = \frac{\log_5 5^3 + \log_5 x}{\log_5 x} = \frac{3 + \log_5 x}{\log_5 x}$. $\log_{25} x = \frac{\log_5 x}{\log_5 25} = \frac{\log_5 x}{\log_5 5^2} = \frac{\log_5 x}{2}$.

Substitute these expressions back into the original equation: $\left(\frac{3 + \log_5 x}{\log_5 x}\right) \cdot \left(\frac{\log_5 x}{2}\right)^2 = 1$.

Let $y = \log_5 x$. Since $x > 0$ and $x \neq 1$, $y$ can be any real number except $\log_5 1 = 0$. The equation becomes: $\left(\frac{3 + y}{y}\right) \cdot \left(\frac{y}{2}\right)^2 = 1$. $\left(\frac{3 + y}{y}\right) \cdot \frac{y^2}{4} = 1$.

Since $y \neq 0$, we can simplify the term $\frac{y^2}{y}$: $\frac{(3 + y) \cdot y}{4} = 1$. Multiply both sides by 4: $(3 + y) y = 4$. $3y + y^2 = 4$. Rearrange the terms to form a quadratic equation: $y^2 + 3y - 4 = 0$.

We can solve this quadratic equation by factoring. We look for two numbers that multiply to -4 and add up to 3. These numbers are 4 and -1. $(y + 4)(y - 1) = 0$.

This gives two possible values for $y$: $y + 4 = 0 \implies y = -4$. $y - 1 = 0 \implies y = 1$.

Now, substitute back $y = \log_5 x$ to find the values of $x$.

Case 1: $y = -4$. $\log_5 x = -4$. By the definition of logarithms, $x = 5^{-4}$. $x = \frac{1}{5^4} = \frac{1}{625}$. This value of $x$ is positive and not equal to 1, so it is a valid solution.

Case 2: $y = 1$. $\log_5 x = 1$. By the definition of logarithms, $x = 5^1$. $x = 5$. This value of $x$ is positive and not equal to 1, so it is a valid solution.

The solutions to the equation are $x = 5$ and $x = \frac{1}{625}$.