Question

Solve the following equation: ${{2}^{x}}+{{2}^{y}}=20,x+y=6$.
(a) x = 2, y = 4
(b) x = 3, y = 3
(c) x = 3, y = 2
(d) x = 0, y = 2

Answer 1

Hint: Eliminates any variable from both the equations using substitution approach to get an equation in one variable. Now, solve it further to get the values of x and y. Use the property of surds wherever required.

Complete step-by-step answer:

Here we have equations given as
${{2}^{x}}+{{2}^{y}}=20$ ………… (i)
x + y = 60 ………………(ii)
We need to determine values of ‘x’ and ‘y’ from the given relations; hence, we can use substitution methods to solve the functions.
Therefore, we can get the value of ‘x’ from equation (ii) by transforming ‘y’ to the other side. Hence, we get
x = 6-y …………….. (iii)
Substituting the value of ‘x’ in equation (i) from equation (iii), hence, we get
${{2}^{\left( 6-y \right)}}+{{2}^{y}}=20$ ……….. (iv)
Now, we can use property of surds that is, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$or vice-versa is also true.
Hence, we can replace the term ${{2}^{6-y}}$ by $\dfrac{{{2}^{6}}}{{{2}^{y}}}$ using the above relation of surds. So, rewriting the equation (iv), we get
$\dfrac{{{2}^{6}}}{{{2}^{y}}}+{{2}^{y}}=20$
Let us suppose ${{2}^{y}}='t'$and simplify the above equation further. Hence, we get
$\dfrac{64}{t}+\dfrac{t}{1}=20$
Now, taking L.C.M, we get
$\dfrac{64+{{t}^{2}}}{t}=\dfrac{20}{1}$
On cross multiplying the above equation, we get
$\begin{align}
  & 64+{{t}^{2}}=20t \\
 & \Rightarrow {{t}^{2}}-20t+64=0...........(v) \\
\end{align}$
Let us find the roots of equation (v) with the factorization method. So, we can split 20 to 16 and 4 such that multiplication of 16 and 4 would be 64 which is equal to the multiplication of constant term and coefficient of $'{{t}^{2}}'$ of the quadratic.
Hence, we get
${{t}^{2}}-16t-4t+64=0$
Taking ‘t’ as common from first two terms and ‘-4’ from last two terms; so we get
T (t - 16) -4 (t - 16) = 0
(t-4) (t-16) = 0
And hence, we can equate t-4 and t-16 to ‘o’ to get the values of ‘t’.
Hence, t=4 and t=16
Now, we supposed the value of t as ${{2}^{y}}$. So, we can get value of ‘y’ as
${{2}^{y}}=4$ or ${{2}^{y}}=16$
Now, we can write 4 as ${{2}^{2}}$and 16 as ${{2}^{4}}$; so, we get values of ‘y’ as
${{2}^{y}}={{2}^{2}}$ or ${{2}^{y}}={{2}^{4}}$
And hence, on comparison of above equation; we get
y=2 or y=4
Now, we can get value of ‘x’ from equation (iii);
So, if y=2, then x=6-2=4
Hence, pair (x, y) can be given as (4, 2)
Similarly, if y = 4 then x = 6 – 4 =2
Hence, pair (x, y) can be given as (2,4)
So, option (a) is the correct answer from the given options.

Note: One can substitute values of ‘x’ or ‘y’ from equation ${{2}^{x}}+{{2}^{y}}=20$ to $x+y=6$as well. We need use logarithm to eliminate ‘x’ from first equation as
${{2}^{x}}=20-20y$
Take log to both sides on base ‘2’
$x{{\log }_{2}}2={{\log }_{2}}\left( 20-2y \right)$
Hence,
$x={{\log }_{2}}\left( 20-2y \right)$
Now, put the value of ‘x’ in the second equation to get the value of ‘y’. This approach is complex but one can use this method as well.
One can go wrong with the property of surds $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. One may write ${{2}^{6-y}}$as ${{2}^{6}}{{.2}^{y}}$ by using $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m+n}}$ , which is wrong, so take care of property of surds as well in these kind of problems.
One can solve the quadratic ${{t}^{2}}-20t+64=0$ using quadratic formula as well which is given as $$
$\dfrac{-B\pm \sqrt{{{B}^{2}}-4AC}}{2A}$ for $A{{x}^{2}}+Bx+C=0$