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

Mar 31, 2016

First, use the log rule $a \log n = \log {a}^{n}$

Explanation:

${\log}_{10} \left({6}^{2}\right) - {\log}_{10} \left({27}^{\frac{1}{3}}\right) - {\log}_{10} \left(x\right) = 0$

Evaluate, using the exponent rule ${a}^{\frac{n}{m}} = \sqrt[m]{{a}^{n}}$

${\log}_{10} \left(36\right) - {\log}_{10} \left(3\right) - {\log}_{10} \left(x\right) = 0$

Now, use the log rule ${\log}_{a} m - {\log}_{a} n = {\log}_{a} \left(\frac{m}{n}\right)$

${\log}_{10} \left(\frac{\frac{36}{3}}{x}\right) = 0$

${\log}_{10} \left(\frac{12}{x}\right) = 0$

Convert to exponential form: ${\log}_{a} n = x \to {a}^{x} = n$

${10}^{0} = \frac{12}{x}$

$1 = \frac{12}{x}$

$x = 12$

Hopefully this helps!