Given $W = 3^{x + y}$ $W=3^(x+y)$ , how do you solve for x?

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Answer 1 · 2016-11-08T01:01:27

$ln W = ln (3^{x + y})$ $lnW = ln(3^(x + y))$

$ln W = (x + y) ln 3$ $lnW = (x + y)ln3$

$ln W = x ln 3 + y ln 3$ $lnW = xln3 + yln3$

$ln W - y ln 3 = x ln 3$ $lnW - yln3 = xln3$

$\frac{ln W - y ln 3}{ln 3} = x$ $(lnW - yln3)/ln3 = x$

$\frac{ln (\frac{W}{3})}{ln 3} = x$ $(ln(W/3))/ln3 = x$

Hopefully this helps!

Answer 2 · 2016-11-08T01:29:14

$x = {log}_{3} W - y = \frac{log W}{log 3} - y$ $x=log_3 W-y=log W/log 3 - y$ .

Inversely,

$x + y = {log}_{3} W$ $x+y=log_3W$

$x = {log}_{3} W - y$ $x=log_3 W-y$ .

Given W=3^(x+y)W=3x+yW=3^(x+y), how do you solve for x?