How do you solve ${log}_{8} 48 - {log}_{8} w = {log}_{8} 6$ $log_8 48-log_8 w=log_8 6$ ?

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How do you solve ${log}_{8} 48 - {log}_{8} w = {log}_{8} 6$ $log_8 48-log_8 w=log_8 6$ ?

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Answer 1 · 2016-10-29T14:07:26

Answer is $w$ $w$ = 8.

Explanation:

${log}_{8} 48 - {log}_{8} w = {log}_{8} 6$ $log_8 48-log_8w=log_8 6$
$:.$ $log48/log8-logw/log8=log6/log8$
$:.$ $log48-logw=log6$
$:.$ $log8*6-logw=log6$
$:.$ $log8+log6-logw=log6$
$:. log8-logw=0$
$:.logw=log8$
$:.w=8$ . (answer).

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How do you solve ${log}_{8} 48 - {log}_{8} w = {log}_{8} 6$ $log_8 48-log_8 w=log_8 6$ ?

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How do you solve log_8 48-log_8 w=log_8 6log848−log8w=log86log_8 48-log_8 w=log_8 6?

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How do you solve ${log}_{8} 48 - {log}_{8} w = {log}_{8} 6$ $log_8 48-log_8 w=log_8 6$ ?