Question

27log(1-4x) = 8x² +1

Answer 1

Відповідь:First, let’s isolate the logarithmic term on one side: [27\log(1-4x) = 8x^2 + 1]

Next, divide both sides of the equation by 27: [\log(1-4x) = \frac{8x^2 + 1}{27}]

Now, let’s get rid of the logarithm by exponentiating both sides with base 10: [1-4x = 10{\frac{8x2 + 1}{27}}]

Rearrange the equation: [10{\frac{8x2 + 1}{27}} - 1 = 4x]

Finally, solve for (x): [x = \frac{10{\frac{8x2 + 1}{27}} - 1}{4}]

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