How do you solve ${log}_{3} 42 - {log}_{3} n = {log}_{3} 7$ $log_3 42-log_3n=log_3 7$ ?

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Answer 1 · 2018-04-30T18:10:49

$n = 6$ $n= 6$

We should use

${log}_{a} n - {log}_{a} m = {log}_{a} (\frac{n}{m})$ $log_a n - log_a m = log_a(n/m)$ .

Therefore:

${log}_{3} (\frac{42}{n}) = {log}_{3} 7$ $log_3(42/n) = log_3 7$

$\frac{42}{n} = 7$ $42/n = 7$

$7 n = 42$ $7n =42$

$n = 6$ $n = 6$

Hopefully this helps!

How do you solve log342−log3n=log37log_3 42-log_3n=log_3 7?