Question

How do you solve ${{6}^{x}}+{{4}^{x}}={{9}^{x}}$?

Answer 1

Hint: The given equation has exponential functions. While solving these types of problems, we should try to convert them into a form of the quadratic equation. To convert the exponentials into variables, we can use natural logarithms $\left( {{e}^{\ln x}}=x \right)$. After getting a quadratic equation, we can easily find out the roots of it.

Complete Step by Step Solution:
The given equation is
$\Rightarrow {{6}^{x}}+{{4}^{x}}={{9}^{x}}$……(1)
We have to convert this equation into a quadratic equation. The standard form of a second-degree quadratic equation is $a{{x}^{2}}+bx+c=0$, where a, b and c are coefficients. Let us try to establish a constant in the given equation in our first step.
If we divide the above equation by ${{4}^{x}}$, we can get a constant in our equation
$\Rightarrow \dfrac{{{6}^{x}}}{{{4}^{x}}}+1=\dfrac{{{9}^{x}}}{{{4}^{x}}}$
We know that $\left( \dfrac{{{x}^{n}}}{{{y}^{n}}} \right)={{\left( \dfrac{x}{y} \right)}^{n}}$. Therefore
$\Rightarrow {{\left( \dfrac{6}{4} \right)}^{x}}+1={{\left( \dfrac{9}{4} \right)}^{x}}$
We also need a component similar to ${{x}^{2}}$
$\Rightarrow {{\left( \dfrac{3}{2} \right)}^{x}}+1={{\left( \dfrac{9}{4} \right)}^{x}}$
$\Rightarrow {{\left( \dfrac{3}{2} \right)}^{x}}+1={{\left( {{\left( \dfrac{3}{2} \right)}^{2}} \right)}^{x}}$
If we assume that $X={{\left( {}^{3}/{}_{2} \right)}^{x}}$, we can rearrange the above equation as shown below.
$\Rightarrow X+1={{X}^{2}}$
$\Rightarrow {{X}^{2}}-X-1=0$ ……(2)
Now that we have a proper second-degree quadratic equation, let us find out its root. We already know that the root of a quadratic equation is equal to $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Let us substitute the coefficients of equation (2) to get the root.
$\Rightarrow X=\dfrac{1\pm \sqrt{1-4\times 1\times \left( -1 \right)}}{2\times 1}$
$\Rightarrow X=\dfrac{1\pm \sqrt{5}}{2}$
 The positive root of the quadratic equation will be equal to
$\Rightarrow X=\dfrac{1+\sqrt{5}}{2}$ ……(3)
Earlier in this problem, we assumed that the variable $X={{\left( {}^{3}/{}_{2} \right)}^{x}}$. Hence the equation (3) can be written as
$\Rightarrow {{\left( \dfrac{3}{2} \right)}^{x}}=\dfrac{1+\sqrt{5}}{2}$ ……(4)
To convert the exponential in the above equation into a variable, let us take natural logarithms on both sides of the equation.
A natural logarithm is nothing but a logarithm with a base equal to the Euler's constant, e. The value e is one of the important mathematical constants like pi $\left( \pi \right)$. It is chosen as a base for numerous reasons. For instance, it is an irrational number having a value that is approximately equal to 2.71828.
The natural logarithm of an unknown value x is, $\ln x={{\log }_{e}}x$. And so ${{e}^{\ln x}}=x$
Let us use the above property of the natural logarithm to find out the variable x.
Taking natural logarithm on both sides of the equation (4), we get
$\Rightarrow \ln {{\left( \dfrac{3}{2} \right)}^{x}}=\ln \left( \dfrac{1+\sqrt{5}}{2} \right)$
$\Rightarrow x.\ln \left( \dfrac{3}{2} \right)=\ln \left( \dfrac{1+\sqrt{5}}{2} \right)$ $\because {{\log }_{a}}{{b}^{x}}=x{{\log }_{a}}b$
Therefore the variable x is given by the equation
$\Rightarrow x=\dfrac{\ln \left( \dfrac{1+\sqrt{5}}{2} \right)}{\ln \left( \dfrac{3}{2} \right)}$
Find out the natural logarithmic values for the values in the above equation and then substitute them.
$\Rightarrow x=\dfrac{0.481211}{0.405465}$

$\Rightarrow x=1.186814$

Note:
It should be noted that a logarithmic function is an inverse of the corresponding exponential function. One can understand and utilize this relationship to solve the problems involving exponentials.