To solve the equation $3 \cdot \log_2(x) - 2 \cdot \log(5x) = 2$, we need to work through the following steps:

Step 1: Convert the logarithms to a common base

Given that the equation mixes base-2 logarithms with base-10 logarithms, it’s usually easiest to convert everything to the same base. Let's convert $\log_2(x)$ to base 10 using the change of base formula:

$\log_2(x) = \frac{\log(x)}{\log(2)}$

So, the equation becomes:

$3 \cdot \frac{\log(x)}{\log(2)} - 2 \cdot \log(5x) = 2$

Step 2: Simplify the equation

Distribute and simplify the terms:

$\frac{3 \cdot \log(x)}{\log(2)} - 2 \cdot \log(5x) = 2$

Step 3: Express $\log(5x)$

Recall that:

$\log(5x) = \log(5) + \log(x)$

Substitute this back into the equation:

$\frac{3 \cdot \log(x)}{\log(2)} - 2 \cdot \left(\log(5) + \log(x)\right) = 2$

Expanding:

$\frac{3 \cdot \log(x)}{\log(2)} - 2 \cdot \log(5) - 2 \cdot \log(x) = 2$

Step 4: Combine like terms

Group the $\log(x)$ terms together:

$\left(\frac{3}{\log(2)} - 2\right) \cdot \log(x) = 2 + 2 \cdot \log(5)$

Step 5: Solve for $\log(x)$

Now isolate $\log(x)$:

$\log(x) = \frac{2 + 2 \cdot \log(5)}{\frac{3}{\log(2)} - 2}$

Step 6: Calculate the value

Substitute $\log(2) \approx 0.3010$ and $\log(5) \approx 0.6990$:

$\log(x) = \frac{2 + 2 \cdot 0.6990}{\frac{3}{0.3010} - 2}$

Calculate:

$\log(x) = \frac{2 + 1.398}{9.9668 - 2}$

$\log(x) = \frac{3.398}{7.9668} \approx 0.4267$

Finally, solve for $x$:

$x = 10^{0.4267} \approx 2.67$

Final Answer:

The solution is approximately $x \approx 2.67$.

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Do you want any details or have any questions?

Here are 5 related questions for further practice:

1. Solve $2 \cdot \log_3(x) + 3 \cdot \log_2(x) = 5$.
2. Solve $\log_4(x) + 2 \cdot \log(x) = 3$.
3. What is the value of $x$ if $5 \cdot \log(x) - \log_5(x^2) = 1$?
4. Solve for $x$ in the equation $\log(x^2) + \log(x) = 3$.
5. Find the value of $x$ in the equation $4 \cdot \log_5(x) - \log_5(x^3) = 1$.

Tip: Always consider converting logarithms to the same base when solving equations that mix different logarithmic bases.