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Answer 1 · 2017-02-10T16:37:16

$1 - ln \sqrt[3]{2} = x$ $1 - lnroot(3)(2) = x$

Use $a^{- n} = \frac{1}{a^{n}}$ $a^-n = 1/a^n$ :

$\frac{1}{e} = 2 e^{3 x - 4}$ $1/e = 2e^(3x - 4)$

Cross multiply:

$1 = 2 e^{3 x - 4} e^{1}$ $1 = 2e^(3x - 4)e^1$

Use $a^{m} \cdot a^{n} = a^{m + n}$ $a^m * a^n = a^(m + n)$

$1 = 2 e^{3 x - 4 + 1}$ $1 = 2e^(3x - 4 + 1)$

$1 = 2 e^{3 x - 3}$ $1 = 2e^(3x - 3)$

$\frac{1}{2} = e^{3 x - 3}$ $1/2 = e^(3x - 3)$

Take the natural logarithm of both sides.

$ln (\frac{1}{2}) = ln (e^{3 x - 3})$ $ln(1/2) = ln(e^(3x- 3))$

Use ${ln a}^{n} = n ln a$ $lna^n = nlna$ .

$ln (\frac{1}{2}) = (3 x - 3) ln e$ $ln(1/2) = (3x- 3)lne$

$ln$ $ln$ and $e$ $e$ are inverses--their product is $1$ $1$

$ln (\frac{1}{2}) = 3 x - 3$ $ln(1/2) = 3x- 3$

$\frac{1}{3} (ln (\frac{1}{2}) + 3) = x$ $1/3(ln(1/2) + 3) = x$

Use $ln (\frac{a}{b}) = ln a - ln b$ $ln(a/b) = lna - lnb$ .

$\frac{1}{3} (ln 1 - ln 2 + 3) = x$ $1/3(ln1 - ln2 + 3) = x$

We know that $ln 1 = 0$ $ln1= 0$ .

$\frac{1}{3} (3 - ln 2) = x$ $1/3(3 - ln2) = x$

$1 - \frac{1}{3} ln 2 = x$ $1 - 1/3ln2 = x$

Use $a ln n = {ln n}^{a}$ $alnn = lnn^a$ .

$1 - {ln 2}^{\frac{1}{3}} = x$ $1 - ln2^(1/3) = x$

$1 - ln \sqrt[3]{2} = x$ $1 - lnroot(3)(2) = x$

Hopefully this helps!