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:

& {{3}^{x}}{{.5}^{y}}=75.....\left( i \right) \\\ & {{3}^{y}}{{.5}^{x}}=45.....\left( ii \right) \\\ \end{aligned}$$ First of all, we will multiply equation (i) and (ii) and get $${{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.