Question

Find x and y if ${{3}^{x}}={{y}^{2}}$ and ${{y}^{x}}=9$

Answer 1

Hint:Now know are given with two-equation, ${{3}^{x}}={{y}^{2}}$ and ${{y}^{x}}=9$ . Now to each equation we will take log on both sides and use the property of log which is $\log {{a}^{n}}$ = n × log a. Hence we will get a simplified equation. From the two equations obtained we will solve them simultaneously to find x and y.

Complete step by step answer:
Now we are given that ${{3}^{x}}={{y}^{2}}$
Let us apply log on both sides of the equation
Hence we get
$\log {{3}^{x}}=\log {{y}^{2}}.............(1)$
Now we have a property in log according to which we have $\log {{a}^{n}}$ = n × log a.
Using this in equation (1) we get
$x\log 3=2\log y...................(2)$
Now taking 2 on the left-hand side we can rewrite equation (2) as
$\log y=\dfrac{x\log 3}{2}.......................(3)$
Now we are given the second equation as ${{y}^{x}}=9$
We know that 9 is square of 9 hence if we write 9 = ${{3}^{2}}$ in the above equation we get
${{y}^{x}}={{3}^{2}}$ .
Now let us again take log on both sides.
$\log {{y}^{x}}=\log {{3}^{2}}.............(4)$
Now again using the property of log in equation (4) which says $\log {{a}^{n}}$ = n log a we get.
$x\log y=2\log 3$.
Now dividing the above equation by x we get
$\log y=\dfrac{2\log 3}{x}..................(5)$
Now LHS in equation (3) and equation (5) are the same hence RHS of the equations also must be the same. Hence equating RHS we get
$\dfrac{2\log 3}{x}=\dfrac{x\log 3}{2}$
Now dividing by log 3 on both sides we get
$\dfrac{2}{x}=\dfrac{x}{2}..............(6)$
Now multiplying equation (6) by 2x on LHS and RHS we get
$\begin{align}
  & \left( 2x \right)\times \dfrac{2}{x}=\left( 2x \right)\times \dfrac{x}{2} \\
 & \Rightarrow 4={{x}^{2}} \\
\end{align}$
Now taking square root on both sides we get
$x=\pm 2$
Now if x = - 2 then we get in equation ${{y}^{x}}=9$ we get
$\begin{align}
  & {{y}^{-2}}=\dfrac{9}{1} \\
 & \Rightarrow {{y}^{2}}=\dfrac{1}{9} \\
 & \Rightarrow y=\pm \dfrac{1}{3} \\
\end{align}$
Now if we check the values in equation ${{3}^{x}}={{y}^{2}}$
For x = -2 we get
$\begin{align}
  & {{3}^{-2}}={{y}^{2}} \\
 & \Rightarrow \dfrac{1}{{{3}^{2}}}={{y}^{2}} \\
 & \Rightarrow y=\pm \dfrac{1}{\sqrt{3}} \\
\end{align}$
Now since the values doesn’t match we will have to discard the possibility of x = -2
Now if we substitute x = 2 in equation ${{y}^{x}}=9$ we get
$\begin{align}
  & {{y}^{2}}=9 \\
 & \Rightarrow y=\pm 3 \\
\end{align}$
And if we substitute x = 2 in ${{3}^{x}}={{y}^{2}}$
We get
$\begin{align}
  & {{3}^{2}}=y \\
 & \Rightarrow {{y}^{2}}=9 \\
 & \Rightarrow y=\pm 3 \\
\end{align}$
Hence the values match. Hence x = 2 and $y=\pm 3$ is the solution.

Note:
 We can also solve this just by looking at the equation since we have ${{3}^{x}}={{y}^{2}}$ we can directly take a guess that the values might be x = 2 and $y=\pm 3$. Now substituting values in ${{y}^{x}}=9$ we can see that x = 2 and $y=\pm 3$are correct answers.