Question

The value of x, which satisfies $4^x+6^x=9^x$ is
A. ${\displaystyle \frac{\ln (\sqrt{5}-1)-\ln 2}{\ln 3-\ln 2}}$
B. ${\displaystyle \frac{\ln (\sqrt{5}-1)+\ln 2}{\ln 2-\ln 3}}$
C. ${\displaystyle \frac{\ln (\sqrt{5}-1)-\ln 2}{\ln 2-\ln 3}}$
D. ${\displaystyle \frac{\ln (\sqrt{5}-1)+\ln 2}{\ln 3-\ln 2}}$

Answer 1

Hint: In this question use the concept of logarithm and also remember the relation between common logarithm and natural logarithm i.e. log and ln which is given as; $\ln x=2.303\log x$ or $\log x={\displaystyle \frac{\ln x}{2.303}}$ and use the formula $\log \left({\displaystyle \frac{a}{b}}\right)=\log a-\log b$ to approach the solution.

Complete step-by-step answer:
According to the given information we have equation $4^x+6^x=9^x$
We have to find the value of x so now $4^x+6^x=9^x$ (equation 1)
Now let’s divide equation (1) by $9^x$ both the sides, we have
 ${\left({\displaystyle \frac{4}{9}}\right)}^x+{\left({\displaystyle \frac{6}{9}}\right)}^x={\displaystyle \frac{9^x}{9^x}}$
 $\Rightarrow$ ${\left({\displaystyle \frac{4}{9}}\right)}^x+{\left({\displaystyle \frac{2}{3}}\right)}^x=1$
We can write the above equation as

 ${\left({\left({\displaystyle \frac{2}{3}}\right)}^2\right)}^x+{\left({\displaystyle \frac{2}{3}}\right)}^x=1$ (equation 2)
Let us substitute ${\left({\displaystyle \frac{2}{3}}\right)}^x=y$ in equation 2
 $y^2+y=1$
 $\Rightarrow$ $y^2+y-1=0$
If we have a quadratic equation of the form $a{x}^2+bx+c=0$
Then the roots of equation of this form is written as $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$ .
Now using the above to solve for the roots of equation $y^2+y-1=0$

 $y={\displaystyle \frac{-1\pm \sqrt{5}}{2}}$
So, we get 2 values of y that is $y={\displaystyle \frac{-1+\sqrt{5}}{2}}$ , $y={\displaystyle \frac{-1-\sqrt{5}}{2}}$
Now we have substituted $y={\left({\displaystyle \frac{2}{3}}\right)}^x$ , so let’s substitute it back
 $y={\left({\displaystyle \frac{2}{3}}\right)}^x={\displaystyle \frac{-1\pm \sqrt{5}}{2}}$
So
 ${\left({\displaystyle \frac{2}{3}}\right)}^x={\displaystyle \frac{-1\pm \sqrt{5}}{2}}$
But ${\left({\displaystyle \frac{2}{3}}\right)}^x={\displaystyle \frac{-1-\sqrt{5}}{2}}$ is not possible since ${\left({\displaystyle \frac{2}{3}}\right)}^x$ will always be positive irrespective of value of x as exponential values aren’t negative.
So only value left with us is ${\left({\displaystyle \frac{2}{3}}\right)}^x={\displaystyle \frac{-1+\sqrt{5}}{2}}$
Now take log both sides, so we get
 $x\log \left({\displaystyle \frac{2}{3}}\right)=\log \left({\displaystyle \frac{\sqrt{5}-1}{2}}\right)$
So, on simplifying
 $x={\displaystyle \frac{\log \left({\displaystyle \frac{\sqrt{5}-1}{2}}\right)}{\log \left({\displaystyle \frac{2}{3}}\right)}}$
Now using the property of logarithm that $\log \left({\displaystyle \frac{a}{b}}\right)=\log a-\log b$
We have
 $x={\displaystyle \frac{\log \left(\sqrt{5}-1\right)-\log 2}{\log 2-\log 3}}$
As we know that $\ln x=2.303\log x$ or $\log x={\displaystyle \frac{\ln x}{2.303}}$
Therefore, $$x={\displaystyle \frac{{\displaystyle \frac{\ln \left(\sqrt{5}-1\right)}{2.303}}-{\displaystyle \frac{\ln 2}{2.303}}}{{\displaystyle \frac{\ln 2}{2.303}}-{\displaystyle \frac{\ln 3}{2.303}}}}$$
After simplifying the above equation, we get
$$x={\displaystyle \frac{\ln \left(\sqrt{5}-1\right)-\ln 2}{\ln 2-\ln 3}}$$
Therefore, the value of x is equal to $${\displaystyle \frac{\ln \left(\sqrt{5}-1\right)-\ln 2}{\ln 2-\ln 3}}$$

So, the correct answer is “Option C”.

Note: Whenever we face such types of problems the key step that we need we have in our mind is that we always try and form a quadratic equation by proper simplification and substitution. This quadratic equation thus formed can be easily solved (roots can be calculated) using the formula of quadratic formula for quadratic equations mentioned above.