A)(1/3)^4-2x=9 B) log0,1(x^2-3x)=-1

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вопрос

A)
First, let's simplify the left side of the equation:
(1/3)^4 = 1/81
(1/3)^4 2x = (1/81) 2x = 2x/81

Now, the equation becomes:
2x/81 = 9

To solve for x, multiply both sides by 81:
2x = 729

Divide by 2:
x = 729/2
x = 364.5

Therefore, the solution to equation A is x = 364.5.

B)
To solve this equation, first isolate the logarithm:
log0.1(x^2 - 3x) = -1

Rewrite the equation in exponential form:
0.1^(-1) = x^2 - 3x

Calculate the value of 0.1^-1:
10 = x^2 - 3x

Rearrange the equation and set it to zero:
x^2 - 3x - 10 = 0

Now, this is a quadratic equation that can be factored or solved using the quadratic formula to find the solutions for x.