Name: Logarithmic Inequalities
Uploaded: 2019-07-19
Duration: 10 min 24 s
Description: Logarithmic inequalities are inequalities in which one (or both) sides involve a logarithm. Watch this video and learn how to solve logarithmic inequalities.

Question

For

a > 0,

solve

2 \log_{x} a + \log_{a x} a + 3 \log_{a^{2} x} a = 0

Accepted Answer

It is given that for $a > 0$

We need to solve $2 \log_{x}^{a} + \log_{a x}^{a} + 3 \log_{a^{2} x}^{a} = 0$

Now the given equation can be rewritten by the property which is $\log_{c}^{d} = \frac{\log d}{\log c}$

$\frac{2 \log a}{\log x} + \frac{\log a}{\log a x} + \frac{3 \log a}{\log a^{2} x} = 0$

Now we take the common term $\log a$ , from the equation, our equation becomes:

$\frac{2}{\log x} + \frac{1}{\log a x} + \frac{3}{\log a^{2} x} = 0$

Now we take L.C.M on the above equation, we get

$\frac{2 \log a x . (\log a^{2} x) + \log x . (\log a^{2} x) + 3 \log x . \log a x}{(\log x) (\log a x) (\log a^{2} x)} = 0$ $2 [\log a x . \log a^{2} x] + \log x . \log a^{2} x + 3 \log x . \log a x = 0$

$2 [(\log a + \log x) . (2 \log a + \log x) + \log x . (2 \log a + \log x) + 3 (\log x . (\log a + \log x)] = 0 (∵ \log (c d) = \log c + \log d)$

$2 [2 {(\log a)}^{2} + {(\log x)}^{2} + 3 \log a . \log x] + 2 \log x . \log a + {(\log x)}^{2} + 3 (\log x . \log a) + 3 {(\log x)}^{2} = 0$

Further, solving we get

${(\log x)}^{2} [2 + 1 + 3] + {(\log a)}^{2} [4] + \log a . \log x [6 + 2 + 3] = 0$

$6 {(\log x)}^{2} + 4 {(\log a)}^{2} + 11 \log x . \log a = 0$

Now we assume that $\log x = t$ , and

$\log a = y$

$6 t^{2} + 4 y^{2} + 11 t y = 0$

Or it can be rewritten as

$6 t^{2} + 11 t y + 4 y^{2} = 0$

Now we factorize the equation:

$6 t^{2} + 3 t y + 8 t y + 4 y^{2} = 0$

$3 t (2 t + y) + 4 y (2 t + y) = 0$

$(2 t + y) (3 t + 4 y) = 0$

$2 t + y = 0 o r 3 t + 4 y = 0$

Further, solving we get,

$2 t = - y o r 3 t = - 4 y$

$t = - \frac{y}{2} o r t = - \frac{4}{3} y$

or

$\log x = - \frac{\log a}{2}, \log x = - \frac{4}{3} \log a$

Canceling $\log$ , from both sides we get

$∴ x = a^{- \frac{4}{3}}$ or

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

Hence, the solution $2 \log_{x} a + \log_{a x} a + 3 \log_{a^{2} x} a = 0$ ,is either $x = a^{- \frac{4}{3}}$ or

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

For $a > 0,$ solve $2 \log_{x} a + \log_{a x} a + 3 \log_{a^{2} x} a = 0$

Important Questions on Appendix II : Logarithm

Learn and Prepare for any exam you want !

For a>0, solve 2logxa+logaxa+3loga2xa=0

Important Questions on Appendix II : Logarithm

Learn and Prepare for any exam you want !

For $a > 0,$ solve $2 \log_{x} a + \log_{a x} a + 3 \log_{a^{2} x} a = 0$