Question #f472a

Question

Question #f472a

1 Answer

Impact of this question

1273 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-09T19:54:36

$z = \frac{y}{3 \frac{y}{x} + \frac{3}{2}}$ $z= y/(3y/x+3/2)$

Explanation:

First, you want to turn the $x$ $x$ , $y$ $y$ and $z$ $z$ powers into factors. To do so, use the naperian logarithm:

$N o t e : ln (a \cdot b) = ln (a) + ln (b) and ln (a^{b}) = b ln (a)$ $color(gray)(Note: ln(a*b) = ln(a)+ln(b) and ln(a^b) = bln(a))$

$ln (2^{x}) = ln (3^{y}) = ln ({(24 \sqrt{3})}^{z})$ $ln(2^x) = ln(3^y) = ln((24sqrt3)^z)$

$= x ln (2) = y ln (3) = z ln (24 \sqrt{3})$ $= xln(2) = yln(3) = zln(24sqrt3)$

Then, isolate $z$ $z$ :

$z = \frac{y ln (3)}{ln (24 \sqrt{3})} = \frac{y ln (3)}{ln (3 \cdot 2^{3}) + ln (3^{\frac{1}{2}})}$ $z = (yln(3))/(ln(24sqrt3)) = (yln(3))/(ln(3*2^3)+ln(3^(1/2)))$

$= \frac{y ln (3)}{3 ln (2) + ln (3) + \frac{ln (3)}{2}}$ $= (yln(3))/(3ln(2)+ln(3)+ln(3)/2)$

Replace $ln (2)$ $ln(2)$ with $\frac{y}{x} ln (3)$ $y/xln(3)$

$z = \frac{y ln (3)}{3 \frac{y}{x} ln (3) + ln (3) + \frac{ln (3)}{2}}$ $z= (yln(3))/(3y/xln(3)+ln(3)+ln(3)/2)$

$= (ycancel(ln(3)))/((3y/x+1+1/2)cancel(ln(3))) = y/(3y/x+3/2)$

Science

Math

Humanities

... and beyond

Question #f472a

1 Answer

Explanation:

Related questions

Impact of this question