Log28 x + log28(x-27)<1

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вопрос

To solve this inequality, we first need to combine the two logarithms using the properties of logarithms.

log28(x) + log28(x-27) < 1

Using the product rule of logarithms, we can combine the two logarithms:

log28(x(x-27)) < 1

Now, we can rewrite the inequality using exponential form:

x(x-27) < 28

Expanding the left side, we get:

x^2 - 27x < 28

Rearranging the inequality, we get:

x^2 - 27x - 28 < 0

Now, we can solve this quadratic inequality by factoring or using the quadratic formula.

Factoring the quadratic, we get:

(x-28)(x+1) < 0

This means that the solutions to the inequality are x < -1 and 28. However, we also need to consider the constraint on the argument of the logarithm:

x > 0
x-27 > 0
x > 27

Combining the two constraints x > 27 and x < 28, we get the final solution:

27 < x < 28