Question

Find value of $x$ if ${4^x} + {6^x} = {9^x}$
A.$\dfrac{{\ln (\sqrt 3 ( -1)) + \ln 2}}{{\ln 2 - \ln 3}}$
B.$\dfrac{{\ln (\sqrt 5 - 1) + \ln 2}}{{\ln 2 - \ln 3}}$
C.$\dfrac{{\ln (\sqrt 5 - 1) - \ln 2}}{{\ln 2 - \ln 3}}$
D.$\dfrac{{\ln (\sqrt 5 - 1) + \ln 2}}{{\ln 3 - \ln 2}}$

Answer 1

To solve this question, firstly we need to simplify the given equation. In this given question, we will convert 4 and 9 numbers to a square of another number and 6 as a product of any two numbers. Then we will substitute these values in the given equation and simplify it. We will then get a quadratic equation and solve it further to get the answer.

Complete step by step solution:
We are given that, ${4^x} + {6^x} = {9^x}$.
Now to simplify the above equation we take 4 as square of 2 and 9 as square of 3 and 6 as a product of 2 and 3.
So, we can rewrite the given equation as
$\Rightarrow {\left( {{2^2}} \right)^x} + {\left( {2 \times 3} \right)^x} = {\left( {{3^2}} \right)^x}$
$\Rightarrow {2^{2x}} + {2^x} \times {3^x} = {3^{2x}}$
Now subtracting ${3^{2x}}$ from both sides, we get
$\Rightarrow {2^{2x}} + {2^x} \times {3^x} - {3^{2x}} = 0$
Now dividing both sides by ${3^{2x}}$, we get
$\Rightarrow \dfrac{{{2^{2x}}}}{{{3^{2x}}}} + \dfrac{{{2^x} \times {3^x}}}{{{3^{2x}}}} - 1 = 0$
Now simplifying the terms, we get
$\Rightarrow {\left( {\dfrac{2}{3}} \right)^{2x}} + {\left( {\dfrac{2}{3}} \right)^x} - 1 = 0$
Now to simplify the above equation we will put ${\left( {\dfrac{2}{3}} \right)^x}$ as $p$.
By substituting this value in the above equation, we get
$\Rightarrow {p^2} + p - 1 = 0$
Now we can see that the above equation is in a quadratic form, so to solve it we can directly use the quadratic formula which is given as, $p = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .
Substituting $a = 1$, $b = 1$ and $c = - 1$in $p = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, we get
$\Rightarrow p = \dfrac{{ - 1 \pm \sqrt {1 - 4\left( 1 \right)\left( { - 1} \right)} }}{2}$
$\Rightarrow p = \dfrac{{ - 1 \pm \sqrt 5 }}{2}$
As $p$ cannot be negative, so we only take its positive value. Therefore, we get
$\Rightarrow p = \dfrac{{ - 1 + \sqrt 5 }}{2}$
$\Rightarrow {\left( {\dfrac{2}{3}} \right)^x} = \dfrac{{ - 1 + \sqrt 5 }}{2}$
Now taking natural logarithm on both the sides, we get
$\Rightarrow \ln {\left( {\dfrac{2}{3}} \right)^x} = \ln \left( {\dfrac{{ - 1 + \sqrt 5 }}{2}} \right)$
Solving the above equation using logarithmic properties, we get
$\Rightarrow x\ln \left( {\dfrac{2}{3}} \right) = \ln \left( { - 1 + \sqrt 5 } \right) - \ln 2$
$\Rightarrow x\left( {\ln 2 - \ln 3} \right) = \ln \left( { - 1 + \sqrt 5 } \right) - \ln 2$
Now dividing both sides by $\left( {\ln 2 - \ln 3} \right)$, we get
$\Rightarrow x = \dfrac{{\ln \left( {\sqrt 5 - 1} \right) - \ln 2}}{{\ln 2 - \ln 3}}$
Therefore, option (C) is the correct option.

Note: Here on solving the given equation, we obtained a quadratic equation. Quadratic equation is a type of equation where the highest degree of the variable is two. That means a quadratic equation has only two roots and not more than that.