Question

Solve the following problems:
\[{{3}^{x}}{{5}^{y}}=75,\text{ }{{3}^{y}}{{5}^{x}}=45\]

Answer 1

Hint: Here, we can multiply and divide the given equations to form the two equations in two given variables, use the rule \[{{a}^{x}}.{{b}^{x}}={{\left( ab \right)}^{x}}\] and \[\dfrac{{{a}^{x}}}{{{b}^{x}}}={{\left( \dfrac{a}{b} \right)}^{x}}\] .

Complete step-by-step answer:

We are given two equations:
\[{{3}^{x}}{{5}^{y}}=75,\text{ }{{3}^{y}}{{5}^{x}}=45\]
Here, we have to find the values of x and y by solving these equations.
Equations are:
\[\begin{align}
  & {{3}^{x}}{{.5}^{y}}=75.....\left( i \right) \\
 & {{3}^{y}}{{.5}^{x}}=45.....\left( ii \right) \\
\end{align}\]
First of all, we will multiply equation (i) and (ii) and get
\[{{3}^{x}}{{.5}^{y}}{{.3}^{y}}{{.5}^{x}}=75\times 45\]
Since, we know that, \[{{a}^{p}}.{{a}^{q}}={{a}^{p+q}}\], we will get,
\[{{3}^{\left( x+y \right)}}{{.5}^{\left( x+y \right)}}=3375\]
Since we know that, \[{{a}^{p}}.{{b}^{p}}={{\left( a.b \right)}^{p}}\], we will get,
\[{{\left( 3.5 \right)}^{\left( x+y \right)}}=3375\]
\[\Rightarrow {{\left( 15 \right)}^{\left( x+y \right)}}=3375\]
Since, we know that, \[3375={{15}^{3}}\], we will get,
\[\Rightarrow {{\left( 15 \right)}^{\left( x+y \right)}}={{\left( 15 \right)}^{3}}\]
We know that when we have, \[{{a}^{p}}={{a}^{q}}\], p and q will be equal for the values when \[a\ne +1,-1,0\]. Using this in the above equation, we can write it as,
\[\left( x+y \right)=3....\left( iii \right)\]
Now, we will divide equation (i) and (ii), we will get,
\[\dfrac{{{3}^{x}}{{.5}^{y}}}{{{3}^{y}}{{.5}^{x}}}=\dfrac{75}{45}\]
Since, we know that \[\dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}}\], we will get,
\[\Rightarrow \dfrac{{{3}^{x-y}}}{{{5}^{x-y}}}=\dfrac{15\times 5}{15\times 3}\]
Since we know that, \[\dfrac{{{a}^{x}}}{{{b}^{x}}}={{\left( \dfrac{a}{b} \right)}^{x}}\], we will get,
\[\Rightarrow {{\left( \dfrac{3}{5} \right)}^{x-y}}=\dfrac{5}{3}\]
Now, we can write \[\dfrac{5}{3}={{\left( \dfrac{3}{5} \right)}^{-1}}\], we will get,
\[{{\left( \dfrac{3}{5} \right)}^{x-y}}={{\left( \dfrac{3}{5} \right)}^{-1}}\]
Therefore, we get,
\[x-y=-1....\left( iv \right)\]
Now by adding equation (iii) and (v), we will get,
\[\left( x+y \right)+\left( x-y \right)=3-1\]
\[\Rightarrow 2x=2\]
Therefore, we get x = 1.
By putting x = 1 in equation (iii), we will get,
\[\left( 1+y \right)=3\]
\[y=3-1\]
Therefore, we get y = 2.

Note: Students can also use the logarithm in the given equations to get the values of x and y in this question. Also, in these types of questions, once you get the value of x and y, always cross-check by substituting the values of x and y back in the equations and see if these values are satisfying the equations or not.