How do you condense $\frac{1}{3} ({log}_{8} y + 2 {log}_{8} (y + 4)) - {log}_{8} (y - 1)$ $1/3(log_8y+2log_8(y+4))-log_8(y-1)$ ?

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Answer 1 · 2017-01-25T21:53:33

${log}_{8} ⎛ ⎜ ⎜ ⎝ \frac{\sqrt[3]{y {(y + 4)}^{2}}}{y - 1} ⎞ ⎟ ⎟ ⎠$ $log_8(root(3)(y(y+4)^2)/(y-1))$

Since $k log a = {log a}^{k}$ $kloga=loga^k$ , the expression is equivalent to:

$\frac{1}{3} ({log}_{8} y + {log}_{8} {(y + 4)}^{2}) - {log}_{8} (y - 1)$ $1/3(log_8y+log_8(y+4)^2)-log_8(y-1)$

Since $log a + log b = log (a b)$ $loga+logb=log(ab)$ , you get:

$\frac{1}{3} {log}_{8} y {(y + 4)}^{2} - {log}_{8} (y - 1)$ $1/3log_8y(y+4)^2-log_8(y-1)$

$= {log}_{8} {(y {(y + 4)}^{2})}^{\frac{1}{3}} - {log}_{8} (y - 1)$ $=log_8(y(y+4)^2)^(1/3)-log_8(y-1)$

Since $log a - log b = log (\frac{a}{b})$ $loga-logb=log(a/b)$ , you get

${log}_{8} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{{(y {(y + 4)}^{2})}^{\frac{1}{3}}}{y - 1} ⎞ ⎟ ⎟ ⎟ ⎠$ $log_8((y(y+4)^2)^(1/3)/(y-1))$

that can be written as:

${log}_{8} ⎛ ⎜ ⎜ ⎝ \frac{\sqrt[3]{y {(y + 4)}^{2}}}{y - 1} ⎞ ⎟ ⎟ ⎠$ $log_8(root(3)(y(y+4)^2)/(y-1))$

How do you condense 1/3(log_8y+2log_8(y+4))-log_8(y-1)13(log8y+2log8(y+4))−log8(y−1)1/3(log_8y+2log_8(y+4))-log_8(y-1)?