Question

How do you solve ${{4}^{x}}+6\left( {{4}^{-x}} \right)=5$?

Answer 1

Hint: In this question, we need to find the value of x from the given equation. We will first use the property of the exponent that ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$ to simplify the equation. Then we will substitute ${{4}^{x}}$ as a and form an equation of degree 2. We will solve this quadratic equation by splitting the middle term method. After that we will get two values of $a={{4}^{x}}$. We will take log with base 4 on both sides and solve to find the value of x. We will use the property of logarithm that ${{\log }_{a}}a=1\text{ and }{{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a$.

Complete step by step answer:
Here, we are given the equation as ${{4}^{x}}+6\left( {{4}^{-x}} \right)=5$. We need to find the value of x. For this, let us first simplify the equation. As we know from the property of exponents that ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$. So we can write ${{4}^{-x}}$ as $\dfrac{1}{{{4}^{x}}}$.
Our equation becomes ${{4}^{x}}+\dfrac{6}{{{4}^{x}}}=5$.
Let us substitute the value of ${{4}^{x}}$ as a to simplify our answer. We have the equation as $a+\dfrac{6}{a}=5$.
Taking LCM of a on both sides we have $\dfrac{{{a}^{2}}+6}{a}=\dfrac{5a}{a}$.
Cancelling a from the denominator on both sides we get ${{a}^{2}}+6=5a$.
Rearranging we get ${{a}^{2}}-5a+6=0$.
Now let us use split the middle term method to solve for the value of a. We need to find two numbers ${{n}_{1}}\text{ and }{{n}_{2}}$ such that ${{n}_{1}}+{{n}_{2}}=-5\text{ and }{{n}_{1}}\cdot {{n}_{2}}=6$. We know that -2-3 = -5 and (-2)(-3) = 6, so ${{n}_{1}}=-2\text{ and }{{n}_{2}}=-3$. Splitting the middle term we get ${{a}^{2}}-2a-3a+6=0$.
Taking a common from the first two terms and -3 common from the last two terms we get $a\left( a-2 \right)-3\left( a-2 \right)=0$.
Taking (a-2) common we get $\left( a-2 \right)\left( a-3 \right)=0$.
Therefore we have a-2 = 0 and a-3 = 0, a = 2 and a = 3.
Splitting the value of a as ${{4}^{x}}$ back, we get ${{4}^{x}}=2\text{ and }{{4}^{x}}=3$.
Let us take log with base 4 on both sides we get ${{\log }_{4}}{{4}^{x}}={{\log }_{4}}2\text{ and }{{\log }_{4}}{{4}^{x}}={{\log }_{4}}3$.
We know that ${{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a$ so let us apply it on the left side of both equation we get $x{{\log }_{4}}4={{\log }_{4}}2\text{ and x}{{\log }_{4}}4={{\log }_{4}}3$.
Now we know that ${{\log }_{b}}b=1$ so using it on the left side of both equation we get $x={{\log }_{4}}2\text{ and x}={{\log }_{4}}3$.
(1) $x={{\log }_{4}}2$.
As we know ${{\left( 2 \right)}^{2}}=4$ and so ${{\left( 4 \right)}^{\dfrac{1}{2}}}=2$. So right hand side can be written as $x={{\log }_{4}}{{4}^{\dfrac{1}{2}}}$.
Using ${{\log }_{b}}{{a}^{m}}=m{{\log }_{b}}a$ we get $x=\dfrac{1}{2}{{\log }_{4}}4$.
Again using ${{\log }_{b}}b=1$ on the right side we get $x=\dfrac{1}{2}$.
(2) $\text{x}{{\log }_{4}}4={{\log }_{4}}3$.
It cannot be solved. So let us convert it into natural log rather than log with base 4. We know ${{\log }_{b}}m=\dfrac{\ln m}{\ln b}$.
Here ln is natural log, so $x=\dfrac{\ln 3}{\ln 4}$.
Hence we get the two values of x as $x=\dfrac{1}{2}$ and $x=\dfrac{\ln 3}{\ln 4}$ which is the required answer.

Note: Students should keep in mind the properties of exponential as well as logarithmic to easily calculate the value of x. Take care of the signs while splitting the middle term. Always try to give answers in natural log or log with base 10.