How do you evaluate ${log}_{49} 7 + {log}_{27} (\frac{1}{9}) \div {log}_{64} (\frac{1}{32}) - {log}_{\frac{3}{2}} (\frac{27}{8})$ $log_49 7 + log_27 (1/9) div log_64 (1/32) - log_(3/2) (27/8)$ ?

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How do you evaluate ${log}_{49} 7 + {log}_{27} (\frac{1}{9}) \div {log}_{64} (\frac{1}{32}) - {log}_{\frac{3}{2}} (\frac{27}{8})$ $log_49 7 + log_27 (1/9) div log_64 (1/32) - log_(3/2) (27/8)$ ?

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Answer 1 · 2016-03-22T19:00:55

${log}_{49} 7 + {log}_{27} (\frac{1}{9}) \div {log}_{64} (\frac{1}{32}) - {log}_{\frac{3}{2}} (\frac{27}{8}) = - \frac{17}{10}$ $log_49 7 +log_27 (1/9) -: log_64 (1/32) - log_(3/2) (27/8) = -17/10$

Explanation:

${log}_{49} 7 + {log}_{27} (\frac{1}{9}) \div {log}_{64} (\frac{1}{32}) - {log}_{\frac{3}{2}} (\frac{27}{8})$ $log_49 7 +log_27 (1/9) -: log_64 (1/32) - log_(3/2) (27/8)$
$= (\frac{log 7}{{log 7}^{2}}) + (\frac{{log 3}^{- 2}}{{log 3}^{3}}) \div (\frac{{log 2}^{- 5}}{{log 2}^{6}}) - ⎛ ⎜ ⎜ ⎝ \frac{{log (\frac{3}{2})}^{3}}{log (\frac{3}{2})} ⎞ ⎟ ⎟ ⎠$ $=(log7/log7^2) +( log 3^-2 /log3^3)-:(log 2^-5 /log 2^6 )-(log (3/2)^3 /log(3/2))$
$= (\frac{log 7}{2 log 7}) + (\frac{- 2 log 3}{3 log 3}) \div (\frac{- 5 log 2}{6 log 2}) - ⎛ ⎜ ⎝ \frac{3 log (\frac{3}{2})}{log (\frac{3}{2})} ⎞ ⎟ ⎠$ $=(log7/(2 log7)) +( (-2 log 3) /(3 log3))-:((-5 log 2) /(6log 2) )-((3log (3/2)) /log(3/2))$
$= \frac{1}{2} - \frac{2}{3} \div - \frac{5}{6} - 3$ $=1/2 -2/3-:-5/6 -3$
$= - \frac{17}{10}$ $=-17/10$

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How do you evaluate ${log}_{49} 7 + {log}_{27} (\frac{1}{9}) \div {log}_{64} (\frac{1}{32}) - {log}_{\frac{3}{2}} (\frac{27}{8})$ $log_49 7 + log_27 (1/9) div log_64 (1/32) - log_(3/2) (27/8)$ ?

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How do you evaluate log497+log27(19)÷log64(132)−log32(278)log_49 7 + log_27 (1/9) div log_64 (1/32) - log_(3/2) (27/8)?

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How do you evaluate ${log}_{49} 7 + {log}_{27} (\frac{1}{9}) \div {log}_{64} (\frac{1}{32}) - {log}_{\frac{3}{2}} (\frac{27}{8})$ $log_49 7 + log_27 (1/9) div log_64 (1/32) - log_(3/2) (27/8)$ ?