Question

Log^2(100x)+log^2(10)=14+log1/x

Answer 1

Ответ:

это логорифмы???????

Answer 2

To solve the equation log^2(100x) + log^2(10) = 14 + log(1/x), we'll first simplify it step by step.

1. Use log properties: log^2(100x) = (log(100x))^2 log^2(10) = (log(10))^2 log(1/x) = -log(x)

So, our equation becomes:

(log(100x))^2 + (log(10))^2 = 14 - log(x)

2. Now, let's simplify further. Remember that log(10) is equal to 1, and log(100x) can be simplified to log(100) + log(x), which is 2 + log(x):

(2 + log(x))^2 + 1 = 14 - log(x)

3. Expand the square of the binomial:

4 + 4*log(x) + (log(x))^2 + 1 = 14 - log(x)

4. Rearrange the terms:

(log(x))^2 + 4*log(x) + 4 - log(x) + 1 - 14 = 0

5. Combine like terms:

(log(x))^2 + 3*log(x) - 9 = 0

Now, we have a quadratic equation in terms of log(x). To solve for log(x), you can use the quadratic formula:

log(x) = [-3 ± √(3^2 - 41(-9))] / (2*1)

Simplify under the square root:

log(x) = [-3 ± √(9 + 36)] / 2

log(x) = [-3 ± √45] / 2

Now, you have two possible solutions for log(x):

1. log(x) = (-3 + √45) / 2
2. log(x) = (-3 - √45) / 2

To find x, you can exponentiate both sides:

1. x = 10^[(−3 + √45) / 2]
2. x = 10^[(−3 - √45) / 2]

These are the two potential solutions for x. Note that the actual values may need to be approximated using a calculator.