Question

6*7^(х*lg2)-4^(x*lg7)=8

Answer 1

Ответ:

См. ниже

Объяснение:

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Answer 2

To solve the equation $6 \times 7^{x \times \log_2{7}} - 4^{x \times \log_7{4}} = 8$, we need to find the value of $x$.

Step 1: Simplify the equation using properties of logarithms and exponentials. First, let's simplify the terms with logarithms:

• $7^{x \times \log_2{7}}$: Since $a^{b \times \log_a{c}} = c^b$, this term simplifies to $7^x$.
• $4^{x \times \log_7{4}}$: Similarly, this term simplifies to $4^x$.

Now the equation becomes:

$6 \times 7^x - 4^x = 8$

Step 2: Rewrite $4^x$ as $2^{2x}$, as $4 = 2^2$:

$6 \times 7^x - 2^{2x} = 8$

Step 3: Notice that both terms have a common base of 2, so we can express $7^x$ in terms of 2 using the property that $a^{\log_a{b}} = b$:

$6 \times 2^{x \times \log_2{7}} - 2^{2x} = 8$

Since $x \times \log_2{7} = \log_2{7^x} = \log_2{2^{(\log_2{7})^x}}$, we can rewrite it as:

$6 \times 2^{\log_2{2^{(\log_2{7})^x}}} - 2^{2x} = 8$

Simplify further:

$6 \times 2^{(\log_2{7})^x} - 2^{2x} = 8$

Step 4: Now, let $y = (\log_2{7})^x$, so the equation becomes:

$6 \times 2^y - 2^{2x} = 8$

Step 5: Re-arrange the equation:

$6 \times 2^y = 2^{2x} + 8$

Step 6: Divide both sides by 2 to make it easier to work with:

$3 \times 2^y = 2^{2x} + 4$

Step 7: Notice that both sides have a power of 2. Let $z = 2^x$, so the equation becomes:

$3z^2 = z^2 + 4$

Step 8: Subtract $z^2$ from both sides:

$2z^2 = 4$

Step 9: Divide both sides by 2:

$z^2 = 2$

Step 10: Take the square root of both sides:

$z = \sqrt{2}$

Step 11: Recall that $z = 2^x$, so:

$2^x = \sqrt{2}$

Step 12: Take the logarithm base 2 of both sides to solve for $x$:

$x = \log_2{\sqrt{2}} = \frac{1}{2}$

So, the solution to the equation is $x = \frac{1$