Question

Помогите сделать!log(x)2-log(4)X=7/6

Answer 1

Logx_2 - 1/2* log2_x = 7/6;
1/ log2_x - 1/2* log2_x = 7/6;
log2_x = t;
1 / t - 1/2  * t - 7/6 = 0;
6 - 3 t^2 - 7 t = 0;
3 t^2 + 7t - 6 = 0;
D = 49 + 72 = 121= 11^2;
t1 = - 6; ⇒log2_x = 6;   x = 2^6 = 64;
t2 = 2/3;  ⇒ log2_x = 2/3; x = 2^(2/3)= 4^(1/3)=корень кубический из 4

Answer 2

I can help you solve logarithmic equations. Here is one possible solution for your equation:

First, we apply the power rule of logarithms to get:

$$\log(x^2) - \log(4) = \log(x) + \log(x) - \log(4)$$

Then, we apply the subtraction rule of logarithms to get:

$$\log\left(\frac{x^2}{4}\right) = 2\log(x) - \log(4)$$

Next, we use the property that $\log(a) = b$ if and only if $a = 10^b$ to rewrite the equation as:

$$\frac{x^2}{4} = 10^{2\log(x) - \log(4)}$$

Now, we simplify the exponent using the properties of logarithms:

$$\frac{x^2}{4} = 10^{\log(x^2) - \log(4)}$$

$$\frac{x^2}{4} = 10^{\log\left(\frac{x^2}{4}\right)}$$

Since the base and the exponent are the same, we can equate the arguments of the logarithms:

$$\frac{x^2}{4} = \frac{x^2}{4}$$

This equation is true for any value of $x$, so we have infinitely many solutions. However, we need to check if any of them make the original equation undefined. We cannot take the logarithm of a negative or zero number, so we must have:

$$x^2 > 0$$

$$x \neq 0$$

$$4 - x > 0$$

$$x < 4$$

Therefore, the solution set is:

$$\{x \in \mathbb{R} \mid 0 < x < 4\}$$

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I hope this helps you understand how to solve logarithmic equations. If you have any other questions, feel free to ask me. I can also generate creative content such as poems, stories, code, and more. Just let me know what you want me to do.

: https://www.symbolab.com/solver/logarithms-calculator