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21 July, 21:24

Solve each inequality for x. (Enter your answers using interval notation.)

(a) ln (x) < 0

(b) ex > 2

Solve each equation for x.

(a) ln (3x - 17) = 5

Express the given quantity as a single logarithm.

ln (a + b) + ln (a - b) - 5 ln (c)

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Answers (1)
  1. 21 July, 21:41
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    (a) ln (x) = 0

    Then 0 < x < 1

    (b) e^x > 2

    Then ln2 < x < ∞

    (a) ln (3x - 17) = 5

    x = 55.1377197

    ln (a + b) + ln (a - b) - 5ln (c)

    = ln[ (a² - b²) / c^5]

    Step-by-step explanation:

    First Part.

    (a) ln (x) < 0

    => x < e^ (0)

    x < 1 ... (1)

    But the logarithm of 0 is 1, and the logarithm of negative numbers are undefined, we can exclude the values of x ≤ 0.

    In fact the values of x that satisfy this inequalities are between 0 and 1.

    Therefore, we write:

    0 < x < 1

    (b) e^x > 2

    This means x > ln2

    and must be finite.

    We write as:

    ln2 < x < ∞

    Second Part.

    (a) ln (3x - 17) = 5

    3x - 17 = e^5

    3x = 17 + e^5

    x = (1/3) (17 + e^5)

    = 55.1377197

    Third Part.

    We need to write

    ln (a + b) + ln (a - b) - 5ln (c)

    as a single logarithm.

    ln (a + b) + ln (a - b) - 5ln (c)

    = ln (a + b) + ln (a - b) - ln (c^5)

    = ln[ (a + b) (a - b) / (c^5) ]

    = ln[ (a² - b²) / c^5]
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