Math Doubts

Solve $2 \log_{x} a$ $+$ $\log_{ax} a$ $+$ $3\log_{a^2x} a$ $=$ $0$ and find the value of $x$

$a$ and $x$ are literals. Three quantities are involved in forming a logarithmic equation and they are expressed as $\log_{x} a$, $\log_{ax} a$ and $\log_{a^2x} a$. The value of the logarithmic equation is zero and it is asked for finding the value of $x$ by solving it.

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

Step: 1

The logarithmic equation contains three logarithmic terms but each term contains three different bases and also having a number a commonly. The logarithmic equation cannot be solved if it is in this form.

However, it is can transformed in another form to make all three logarithm terms to have same base. Use change of base rule of logarithm to do it.

$\implies$ $2 \times \dfrac{1}{\log_{a} x}$ $+$ $\dfrac{1}{\log_{a} ax}$ $+$ $3 \times \dfrac{1}{\log_{a} a^2x}$ $=$ $0$

$\implies$ $\dfrac{2}{\log_{a} x}$ $+$ $\dfrac{1}{\log_{a} ax}$ $+$ $\dfrac{3}{\log_{a} a^2x}$ $=$ $0$

Step: 2

The denominators of second and third terms of terms of this equation expresses logarithm of product of two numbers. It can be expanded by using the product rule of logarithms.

$\implies$ $\dfrac{2}{\log_{a} x}$ $+$ $\dfrac{1}{\log_{a} a + \log_{a} x}$ $+$ $\dfrac{3}{\log_{a} a^2 + \log_{a} x}$ $=$ $0$

Step: 3

There is a logarithmic term in the denominator of the second term of this equation and it is the logarithm of the base. Its value is one as per log of base rule.

There is a logarithmic term in the denominator of the third term and it contains logarithm of an exponential term. It can be simplified by applying the power rule of the logarithm.

$\implies$ $\dfrac{2}{\log_{a} x}$ $+$ $\dfrac{1}{1+\log_{a} x}$ $+$ $\dfrac{3}{2 \log_{a} a + \log_{a} x}$ $=$ $0$

Once again use the logarithm of base rule to simplify the denominator of the third logarithmic term.

$\implies$ $\dfrac{2}{\log_{a} x}$ $+$ $\dfrac{1}{1+\log_{a} x}$ $+$ $\dfrac{3}{2 \times 1 + \log_{a} x}$ $=$ $0$

$\implies$ $\dfrac{2}{\log_{a} x}$ $+$ $\dfrac{1}{1+\log_{a} x}$ $+$ $\dfrac{3}{2+\log_{a} x}$ $=$ $0$

Step: 4

The equation is successfully transformed as a perfect and simple form. Take $y = \log_{a} x$ and transform the entire equation in terms of $y$ to simplify it easily.

$\implies$ $\dfrac{2}{y}$ $+$ $\dfrac{1}{1+y}$ $+$ $\dfrac{3}{2+y}$ $=$ $0$

$\implies$ $\dfrac{2(1+y)(2+y)+y(2+y)+3y(1+y)}{y(1+y)(2+y)}$ $=$ $0$

$\implies$ $2(1+y)(2+y)$ $+$ $y(2+y)$ $+$ $3y(1+y)$ $=$ $0$

$\implies$ $2(2+y+2y+y^2)$ $+$ $2y+y^2+3y+3y^2$ $=$ $0$

$\implies$ $2(2+3y+y^2)$ $+$ $2y+y^2+3y+3y^2$ $=$ $0$

$\implies$ $4+6y+2y^2+2y+y^2+3y+3y^2$ $=$ $0$

$\implies$ $6y^2+11y+4 = 0$

Step: 5

It is a quadratic equation and it can be solved by the quadratic formula method.

$\implies$ $y = \dfrac{-11 \pm \sqrt{11^2 -4 \times 6 \times 4}}{2 \times 6}$

$\implies$ $y = \dfrac{-11 \pm \sqrt{121-96}}{12}$

$\implies$ $y = \dfrac{-11 \pm \sqrt{25}}{12}$

$\implies$ $y = \dfrac{-11 \pm \sqrt{25}}{12}$

$\implies$ $y = \dfrac{-11 \pm 5}{12}$

$\implies$ $y = \dfrac{-11+5}{12}$ and $y = \dfrac{-11-5}{12}$

$\implies$ $y = \dfrac{-6}{12}$ and $y = \dfrac{-16}{12}$

$\implies$ $\require{cancel} y = \dfrac{\cancel{-6}}{\cancel{12}}$ and $\require{cancel} y = \dfrac{\cancel{-16}}{\cancel{12}}$

$\implies$ $y = -\dfrac{1}{2}$ and $y = -\dfrac{4}{3}$

Step: 6

Now, replace the value of $y$.

$\log_{a} x = -\dfrac{1}{2}$ and $\log_{a} x = -\dfrac{4}{3}$

$x = a^{-\frac{1}{2}}$ and $x = a^{-\frac{4}{3}}$

The literal $x$ has two solutions for this logarithmic equation and it is the required solution for this logarithm maths problem.



Follow us
Email subscription
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more