How do you solve $2log_10 6 - 1/3 log_10 27 - log_10 x = 0$ ?

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Answer 1 · 2016-03-31T23:03:29

First, use the log rule $alogn = loga^n$

$log_10(6^2) - log_10(27^(1/3)) - log_10(x) = 0$

Evaluate, using the exponent rule $a^(n/m) = root(m)(a^n)$

$log_10(36) - log_10(3) - log_10(x) = 0$

Now, use the log rule $log_am - log_an = log_a(m/n)$

$log_10((36/3)/x) = 0$

$log_10(12/x) = 0$

Convert to exponential form: $log_an = x -> a^x = n$

$10^0 = 12/x$

$1 = 12/x$

$x = 12$

Hopefully this helps!

How do you solve 2log_10 6 - 1/3 log_10 27 - log_10 x = 02log_10 6 - 1/3 log_10 27 - log_10 x = 0?