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30 December, 16:07

Solve the equation. Check your answer. If necessary, round to 3 decimal places

In (t - 1) + In x^2 = 6, X=

Algebra 2

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  1. 30 December, 17:51
    0
    x ≈ ±20.086/√ (t - 1)

    Step-by-step explanation:

    ln (t - 1) + ln (x²) = 6

    Recall that lnu + lnv = ln (uv). Then

    ln (t - 1) + ln (x²) = ln[ (t-1) x²] = 6

    Take the natural antilogarithm of each side

    (t - 1) x² = e⁶

    Divide each side by t - 1

    x² = e⁶ / (t-1)

    Take the square root of each side

    x = ±e³/√ (t - 1)

    x ≈ ±20.086/√ (t - 1)
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