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# How do you solve \log _ { 27} ( 2x + 4) + \log _ { 81} ( 6x + 12) = \frac { 5} { 6}?

Mar 8, 2018

$x = - \frac{1}{2}$

#### Explanation:

Using $\log a b = \log \frac{b}{\log} a$, we can simplify given equation and solve it as follows:

${\log}_{27} \left(2 x + 4\right) + {\log}_{81} \left(6 x + 12\right) = \frac{5}{6}$

i.e. $\log \frac{2 x + 4}{\log} 27 + \log \frac{6 x + 12}{\log} 81 = \frac{5}{6}$

or $\log \frac{2 x + 4}{3 \log 3} + \log \frac{6 x + 12}{4 \log 3} = \frac{5}{6}$

or $\log \frac{2 x + 4}{3} + \log \frac{6 x + 12}{4} = \frac{5}{6} \log 3$

or $4 \log \left(2 x + 4\right) + 3 \log \left(6 x + 12\right) = 10 \log 3$

or $\log \left[{\left(2 x + 4\right)}^{4} {\left(6 x + 12\right)}^{3}\right] = \log {3}^{10}$

or ${\left(2 x + 4\right)}^{4} {\left(6 x + 12\right)}^{3} = {3}^{10}$

or ${2}^{4} {\left(x + 2\right)}^{4} \cdot {6}^{3} {\left(x + 2\right)}^{3} = {3}^{10}$

or ${\left(x + 2\right)}^{7} = {3}^{10} / \left({2}^{4} \cdot {6}^{3}\right)$

or ${\left(x + 2\right)}^{7} = {3}^{10} / \left({2}^{4} \cdot {2}^{3} \cdot {3}^{3}\right)$

or ${\left(x + 2\right)}^{7} = {3}^{7} / {2}^{7}$

or $x + 2 = \frac{3}{2}$

or $x = \frac{3}{2} - 2 = - \frac{1}{2}$