Question

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

Answer 1

Hint: In this question, we need to solve the given equation to find the value of the unknown variable x. We will convert the given equation to a quadratic equation to simplify it. Firstly, we will multiply the given equation by ${4^x}$ and obtain a quadratic equation. To make the process easier, we take $t = {4^x}$ and obtain the roots using the quadratic formula. After that we substitute $t = {4^x}$ and obtain the value for ${4^x}$. We then simplify it and obtain the value for the variable x.

Complete step by step solution:
Given the equation of the form ${4^x} + 6({4^{ - x}}) = 5$ …… (1)
We are asked to solve the above equation. i.e. to find the value for the variable x.
Firstly, we make rearrangements in the expression given in the equation (1) to simplify it.
In the expression above, we rewrite ${4^{ - x}} = \dfrac{1}{{{4^x}}}$.
Hence equation (1) can be written as,
$ \Rightarrow {4^x} + 6 \cdot \dfrac{1}{{{4^x}}} = 5$
Now multiplying throughout by ${4^x}$ we get,
$ \Rightarrow {4^x} \cdot {4^x} + 6 \cdot \dfrac{1}{{{4^x}}} \cdot {4^x} = 5 \cdot {4^x}$
$ \Rightarrow {4^{2x}} + 6 = 5 \cdot {4^x}$
Transferring all the terms to the L.H.S. we get,
$ \Rightarrow {4^{2x}} - 5 \cdot {4^x} + 6 = 0$
This can also be written as,
$ \Rightarrow {({4^x})^2} - 5 \cdot {4^x} + 6 = 0$
Now we take $t = {4^x}$. So that it will be easier to solve. Hence we have,
$ \Rightarrow {t^2} - 5t + 6 = 0$ …… (2)
Note that the above equation is a quadratic equation.
If $a{x^2} + bx + c = 0$ is a quadratic equation, then we find the roots using quadratic formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Note that here $a = 1$, $b = - 5$ and $c = 6$.
Hence we get,
$t = \dfrac{{ - ( - 5) \pm \sqrt {{{( - 5)}^2} - 4 \times 1 \times 6} }}{{2 \times 1}}$
$ \Rightarrow t = \dfrac{{5 \pm \sqrt {25 - 24} }}{2}$
$ \Rightarrow t = \dfrac{{5 \pm \sqrt 1 }}{2}$
$ \Rightarrow t = \dfrac{{5 \pm 1}}{2}$
Now substituting back $t = {4^x}$ we get,
$ \Rightarrow {4^x} = \dfrac{{5 \pm 1}}{2}$
When ${4^x} = \dfrac{{5 + 1}}{2}$
Now to isolate ${4^x}$, we take $\ln $ (natural logarithm) on both sides of the above equation we get,
$\ln {4^x} = \ln \left( {\dfrac{{5 + 1}}{2}} \right)$
$ \Rightarrow x\ln 4 = \ln \left( {\dfrac{6}{2}} \right)$ $(\because \ln {a^b} = b\ln a)$
$ \Rightarrow x\ln 4 = \ln 3$
Taking $\ln 4$ to the other side we get,
$ \Rightarrow x = \dfrac{{\ln 3}}{{\ln 4}}$
When ${4^x} = \dfrac{{5 - 1}}{2}$
$ \Rightarrow {4^x} = \dfrac{4}{2}$
$ \Rightarrow {4^x} = 2$
We can write 4 as, $4 = {2^2}$
Hence we get,
$ \Rightarrow {({2^2})^x} = 2$
$ \Rightarrow {2^{2x}} = {2^1}$
Now the base on both sides of the equation is the same which is 2.
We have the result that if the expression having the same bases is equal only if the exponents or indices are also equal.
Hence equating the indices, we get,
$ \Rightarrow 2x = 1$
Taking 2 to the other side we get,
$ \Rightarrow x = \dfrac{1}{2}$
Thus, we have $x = \dfrac{{\ln 3}}{4}$ and $x = \dfrac{1}{2}$

Hence the solution for the equation ${4^x} + 6({4^{ - x}}) = 5$ is given by $x = \dfrac{{\ln 3}}{4}$ and $x = \dfrac{1}{2}$.

Note :
It is important to substitute ${4^x}$as $t = {4^x}$, so that it will be easier to solve the problem.
Consider the general quadratic equation $a{x^2} + bx + c = 0$, then we find the roots using quadratic formula given by $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Students must remember the rules of exponents to simplify such problems. We need to be careful while applying the rules. It is necessary to use the correct rule to split the terms and simplify the answer.
The rules of exponents are given below.
(1) Multiplication rule : ${a^m} \cdot {a^n} = {a^{m + n}}$
(2) Division rule : $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
(3) Power of a power rule : ${({a^m})^n} = {a^{mn}}$
(4) Power of a product rule : ${(ab)^m} = {a^m}{b^m}$
(5) Power of a fraction rule : ${\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}$
(6) Zero exponent : ${a^0} = 1$
(7) Negative exponent : ${a^{ - x}} = \dfrac{1}{{{a^x}}}$