# How do you solve for x in  1/log_3x + 1/log _27 x = 4?

Jul 28, 2018

$x = 3$

#### Explanation:

Just a simple thing to note: $\frac{1}{{\log}_{a} \left(b\right)} = {\log}_{b} \left(a\right)$

Therefore,

$\frac{1}{{\log}_{3} \left(x\right)} + \frac{1}{{\log}_{27} \left(x\right)} = 4$

$\implies {\log}_{x} \left(3\right) + {\log}_{x} \left(27\right) = 4$

Using the fact that ${\log}_{a} \left(b\right) + {\log}_{a} \left(c\right) = {\log}_{a} \left(b c\right)$,

$\implies {\log}_{x} \left(3 \cdot 27\right) = 4$

$\implies {\log}_{x} \left(81\right) = 4$

Since ${\log}_{a} \left(b\right) = c$ is just ${a}^{c} = b ,$

${x}^{4} = 81$ We can solve this equation!

$\implies x = \sqrt{81}$

$\implies x = 3$ or $x = - 3$
For now, you can just ignore the negative answer.

$x = 3$

$x = 3$

#### Explanation:

Given equation:

$\frac{1}{\setminus} {\log}_{3} x + \frac{1}{\setminus} {\log}_{27} x = 4$

$\frac{1}{\setminus} {\log}_{3} x + \frac{1}{\setminus} {\log}_{{3}^{3}} x = 4$

$\frac{1}{\setminus} {\log}_{3} x + \frac{1}{\frac{1}{3} \setminus {\log}_{3} x} = 4 \setminus \quad \left(\setminus \because \setminus \setminus {\log}_{{a}^{n}} b = \frac{1}{n} \setminus {\log}_{a} b\right)$

$\frac{1}{\setminus} {\log}_{3} x + \frac{3}{\setminus {\log}_{3} x} = 4$

$\frac{4}{\setminus {\log}_{3} x} = 4$

$\setminus {\log}_{3} x = \frac{4}{4}$

$\setminus {\log}_{3} x = 1$

$x = 3$