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

Hint: 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.