How do you condense $3 {log}_{2} t - \frac{1}{3} {log}_{2} u + 4 {log}_{2} v$ $3 log_2 t -1/3 log_2 u+4 log_2 v$ ?

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How do you condense $3 {log}_{2} t - \frac{1}{3} {log}_{2} u + 4 {log}_{2} v$ $3 log_2 t -1/3 log_2 u+4 log_2 v$ ?

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Answer 1 · 2016-03-23T10:15:02

${log}_{2} (\frac{t^{3} \cdot v^{4}}{\sqrt[3]{u}})$ $log_2((t^3*v^4)/root(3)u)$

Explanation:

As bases are common and we can condense

$3 {log}_{2} t - \frac{1}{3} {log}_{2} u + 4 {log}_{2} v$ $3log_2t-1/3log_2u+4log_2v$ , using following identities

$a {log}_{x} b = {log}_{x} b^{a}$ $alog_xb=log_xb^a$ , $\frac{1}{a} {log}_{x} b = {log}_{x} b^{\frac{1}{a}} = {log}_{x} \sqrt[a]{b}$ $1/alog_xb=log_xb^(1/a)=log_xroot(a)b$

${log}_{x} p + {log}_{x} q = {log}_{x} p q$ $log_xp+log_xq=log_xpq$ and

${log}_{x} p - {log}_{x} q = {log}_{x} (\frac{p}{q})$ $log_xp-log_xq=log_x(p/q)$ .

Hence, $3 {log}_{2} t - \frac{1}{3} {log}_{2} u + 4 {log}_{2} v$ $3log_2t-1/3log_2u+4log_2v$

= ${log}_{2} t^{3} - {log}_{2} \sqrt[3]{u} + {log}_{2} v^{4}$ $log_2t^3-log_2root(3)u+log_2v^4$

= ${log}_{2} (\frac{t^{3} \cdot v^{4}}{\sqrt[3]{u}})$ $log_2((t^3*v^4)/root(3)u)$

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How do you condense $3 {log}_{2} t - \frac{1}{3} {log}_{2} u + 4 {log}_{2} v$ $3 log_2 t -1/3 log_2 u+4 log_2 v$ ?

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Explanation:

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How do you condense $3 {log}_{2} t - \frac{1}{3} {log}_{2} u + 4 {log}_{2} v$ $3 log_2 t -1/3 log_2 u+4 log_2 v$ ?