Calculus Problem: 11^(x+5)=8e^(8-x)?

Can someone explain how to do this problem?

Comments

  • 11^(x+5) = 8e^(8-x)

    ln 11^(x+5) = ln 8e^(8-x)

    ln 11^(x+5) = ln 8 + ln e^(8-x)

    (x+5) ln 11 = ln 8 + (8-x) ln e

    (x+5) ln 11 = ln 8 + (8-x)

    x ln 11 + x = 8 + ln 8 - 5 ln 11

    x(ln11 + 1) = 8 + ln 8 - 5 ln 11

    x = (8 + ln 8 - 5 ln 11) / (ln11 + 1)

    x ≈ -0.5621

  • 11^(x + 5) = 8e^(8 - x)

    11^(x + 5) = 8•e^(8 - x)

    => Apply the natural logarithmic function "In" to both sides, i.e.

    ln 11^(x + 5) = ln [8•e^(8 - x)]

    => Consider these rules: In a^b = bIn a & In (a•b) = In a + In b, then by applying them, you'll obtain:

    (x + 5) ln 11 = ln 8 + In e^(8 - x)

    => Also, consider this: In e^a = a, then

    xIn 11 + 5In 11 = ln 8 + (8 - x)

    xIn 11 + 5In 11 = In 8 + 8 - x

    => Pair like terms, i.e.

    xln 11 + x = 8 + ln 8 - 5ln 11

    => Factor out x at the left hand side, i.e.

    x(ln 11 + 1) = 8 + ln 8 - 5 ln 11

    => Divide both sides by (In 11 + 1), then

    x = (8 + ln 8 - 5ln 11)/(ln 11 + 1)

    => Using a scientific calculator, compute the expression at the right hand side on the calculator, then

    x ≈ - 0.5621 ...Ans.

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