Question #db5cd

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Answer 1 · 2017-02-10T16:37:16

$1 - lnroot(3)(2) = x$

Use $a^-n = 1/a^n$ :

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

Cross multiply:

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

Use $a^m * a^n = a^(m + n)$

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

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

$1/2 = e^(3x - 3)$

Take the natural logarithm of both sides.

$ln(1/2) = ln(e^(3x- 3))$

Use $lna^n = nlna$ .

$ln(1/2) = (3x- 3)lne$

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

$ln(1/2) = 3x- 3$

$1/3(ln(1/2) + 3) = x$

Use $ln(a/b) = lna - lnb$ .

$1/3(ln1 - ln2 + 3) = x$

We know that $ln1= 0$ .

$1/3(3 - ln2) = x$

$1 - 1/3ln2 = x$

Use $alnn = lnn^a$ .

$1 - ln2^(1/3) = x$

$1 - lnroot(3)(2) = x$

Hopefully this helps!