(3/7)^3x+1=(7/3)^5x-3

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To solve this exponential equation, we can first rewrite the fractions as powers of 7 and 3.

(3/7)^3x+1 = (7/3)^5x-3
(7^(-1))^3x+1 = (3^(-1))^5x-3
7^(-3x-1) = 3^(-5x+3)

Now, we can express both sides of the equation with the same base (either 3 or 7) in order to solve for x.

7^(-3x-1) = 3^(-5x+3)
(7^(-1))^(3x+1) = (3^(-1))^(5x-3)
(1/7)^(3x+1) = (1/3)^(5x-3)

Now, we know that for any positive number a, (1/a) = a^(-1), thus:

1/(7^(3x+1)) = 1/(3^(5x-3))
7^(3x+1) = 3^(5x-3)

Now we have both sides with the same base, so we can solve for x:

(7^3x) 7 = (3^5x) 3^(-3)
7^(3x+1) = 3^(5x) / 3^3
7^(3x+1) = 3^(5x-3)

Now, we can set the exponents equal to each other:

3x + 1 = 5x - 3
3x - 5x = -3 - 1
-2x = -4
x = 2

Therefore, x = 2 is the solution to the equation.