How do you solve log_3x-2log_x3=1?

Mar 6, 2018

See the subscript

Explanation:

Mar 6, 2018

$x = \frac{1}{3} , 9$

Explanation:

${\log}_{b} a = \frac{1}{\log} _ a b$

log_3x−2log_x3=1

log_3x−2/log_3x=1

Let log_3x = a

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

${a}^{2} - a - 2 = 0$

$\left(a - 2\right) \left(a + 1\right) = 0$

$a = - 1 , a = 2$

=> ${\log}_{3} x = - 1 \mathmr{and} {\log}_{3} x = 2$

${\log}_{3} x = - 1 \implies x = {3}^{-} 1 = \frac{1}{3}$
or
${\log}_{3} x = 2 \implies x = {3}^{2} = 9$

$x = \frac{1}{3} , 9$

Mar 6, 2018

$x = 9 \mathmr{and} x = \frac{1}{3}$

Explanation:

Note that $2 {\log}_{x} \left(3\right) \to {\log}_{x} \left({3}^{2}\right) = {\log}_{x} \left(9\right)$

${\log}_{3} \left(x\right) - {\log}_{x} \left(9\right) = 1$

${\log}_{3} \left(x\right) = 1 + {\log}_{x} \left(9\right)$

Consider the LHS $\to {\log}_{10} \frac{x}{\log} _ 10 \left(3\right) \text{ } \ldots E q u a t i o n \left(1\right) = {y}_{1}$

Consider the RHS $\to 1 + {\log}_{10} \frac{9}{{\log}_{10} \left(x\right)} \text{ } . . E q u a t i o n \left(2\right) = {y}_{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As we have got all ${\log}_{10}$ now we can drop the base 10 and use just log.

Try ${y}_{1} = {y}_{2}$ and see what happens.

$\log \frac{x}{\log} \left(3\right) = 1 + \log \frac{9}{\log \left(x\right)}$

$\log \frac{x}{\log} \left(3\right) = \frac{\log \left(x\right) + \log \left(9\right)}{\log} \left(x\right)$

Cross multiply

$\log {\left(x\right)}^{2} = \log \left(x\right) \log \left(3\right) + \log \left(3\right) \log \left(9\right)$

Set $\log \left(x\right) = t$ giving

${t}^{2} - t \log \left(3\right) - \log \left(3\right) \log \left(9\right) = 0 \leftarrow \text{ A quadratic in } t$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$y = a {x}^{2} + b x + c \to x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Where a=1; b=-log(3); c= -log(3)log(9)

$t = \frac{+ \log \left(3\right) \pm \sqrt{\log {\left(3\right)}^{2} - 4 \left(1\right) \left(- \log \left(3\right) \log \left(9\right)\right)}}{2 \left(1\right)}$

$t = + \log \frac{3}{2} \pm \frac{\sqrt{2.0488022 \ldots}}{2}$

$t \approx + 0.95424 \ldots \implies {\log}^{-} 1 \left(t\right) = 9 = x$

$t \approx - 0.4771212 \ldots . \implies {\log}^{- 1} \left(t\right) = \frac{1}{3} = x$