Question

5^ |x-1|=7 решить показательное уравнение

Answer 1

Ответ прекрепила во вложении

Answer 2

To solve the exponential equation 5^|x-1| = 7, we need to eliminate the absolute value first. Remember that the absolute value of a number can be positive or negative, so we will consider both cases.

Case 1: x - 1 is positive (x - 1 > 0) When x - 1 is positive, the absolute value |x - 1| can be removed, and we have: 5^(x - 1) = 7

Case 2: x - 1 is negative (x - 1 < 0) When x - 1 is negative, the absolute value |x - 1| will make it positive, so we need to negate it: 5^(-(x - 1)) = 7

Now, we will solve both equations:

Case 1: 5^(x - 1) = 7 To solve this equation, we can take the logarithm of both sides (with any base you prefer, but the natural logarithm ln is commonly used): ln(5^(x - 1)) = ln(7)

Now, use the logarithm property that ln(a^b) = b * ln(a): (x - 1) * ln(5) = ln(7)

Now, isolate x: x = (ln(7)) / ln(5) + 1

Use a calculator to compute the numerical value.

Case 2: 5^(-(x - 1)) = 7 To solve this equation, we can again take the logarithm of both sides (with any base you prefer): ln(5^(-(x - 1))) = ln(7)

Using the logarithm property that ln(a^b) = b * ln(a): (-(x - 1)) * ln(5) = ln(7)

Now, isolate x: -x + 1 = (ln(7)) / ln(5)

Then, isolate x on one side: x = 1 - (ln(7)) / ln(5)

Use a calculator to compute the numerical value.

These are the solutions for the exponential equation 5^|x-1| = 7, considering both cases. Remember that logarithms are multivalued, so there might be other solutions in the complex plane, but the above calculations provide the real solutions for the equation.