The following two algebraic equations are defined in terms of two variables $x$ and $y$ in this problem.
$(1).\,\,\,$ $25^{\,\displaystyle x\small-1}$ $\,=\,$ $5^{\displaystyle y\small +1}$
$(2).\,\,\,$ $\Bigg(\dfrac{1}{9}\Bigg)^{\displaystyle \small 2\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
The two equations are in exponential form. We have to solve both equations in exponential form for evaluating the variables $x$ and $y$.
Let’s focus on the first equation. The equation is in exponential form completely. The bases of the both sides of exponential expressions are numbers but their exponents are algebraic expressions.
$25^{\,\displaystyle x\small-1}$ $\,=\,$ $5^{\displaystyle y\small +1}$
The bases of the left and right hand side exponential expressions are $25$ and $5$ respectively. The number $25$ can be split as the factors of $5$. It is a step, which allows us to initiate the process of simplifying this exponential equation.
$\implies$ $\Big(5 \times 5\Big)^{\,\displaystyle x\small-1}$ $\,=\,$ $5^{\,\displaystyle y\small+1}$
According to the exponentiation, the product form can be expressed in exponential notation.
$\implies$ $\Big(5^2\Big)^{\,\displaystyle x\small-1}$ $\,=\,$ $5^{\,\displaystyle y\small+1}$
The left hand side exponential expression can be simplified by the power rule of exponents.
$\implies$ $5^{\,\displaystyle \small 2 \normalsize (x\small-1)}$ $\,=\,$ $5^{\,\displaystyle y\small+1}$
In the exponent position of the left hand side exponential equation, the number $2$ can be distributed to difference of the terms by the distributive property.
$\implies$ $5^{\,\displaystyle \small 2 \normalsize \times x\small-2 \times 1}$ $\,=\,$ $5^{\,\displaystyle y\small+1}$
$\implies$ $5^{\,\displaystyle \small 2 \normalsize x\small-2}$ $\,=\,$ $5^{\,\displaystyle y\small+1}$
The bases of the both sides of the exponential equations are equal. Hence, the exponents should be equal mathematically.
$\implies$ $2x-2 \,=\, y+1$
Now, simplify this algebraic equation for solving it in mathematics.
$\implies$ $2x-2-y-1 \,=\, 0$
$\implies$ $2x-y-2-1 \,=\, 0$
$\,\,\,\therefore\,\,\,\,\,\,$ $2x-y-3 \,=\, 0$
This algebraic equation contains two variables and it represents a linear equation in two variables.
It is time to solve the given second exponential equation. This equation is also defined in exponential form and it is similar to the first equation. So, we can simplify this equation by following the same procedure.
$\Bigg(\dfrac{1}{9}\Bigg)^{\displaystyle \small 2\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
Look at the bases of the exponential expressions in the both sides of the equation. The bases are rational numbers. The numbers in the numerators are the same but their denominators are different. However, the number $9$ can be factored in terms of $3$.
$\implies$ $\Bigg(\dfrac{1}{3 \times 3}\Bigg)^{\displaystyle \small 2\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
$\implies$ $\Bigg(\dfrac{1}{3^2}\Bigg)^{\displaystyle \small 2\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
$\implies$ $\Bigg(\dfrac{1^2}{3^2}\Bigg)^{\displaystyle \small 2\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
The exponents of both numerator and denominator are the same in the base of the left hand expression. Hence we can simplify it by the power of a quotient rule.
$\implies$ $\Bigg(\Big(\dfrac{1}{3}\Big)^2\Bigg)^{\displaystyle \small 2\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
Now, use the power rule of an exponential function for simplifying this exponential equation further.
$\implies$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 2 \times 2\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
$\implies$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 4\normalsize y}$ $\,=\,$ $\Bigg(\dfrac{1}{3}\Bigg)^{\displaystyle \small 3-\normalsize x}$
We know that the exponents should be equal when their bases are equal.
$\implies$ $4y \,=\, 3-x$
$\,\,\,\therefore\,\,\,\,\,\,$ $x+4y-3 \,=\, 0$
This equation is also a linear equation in two variables.
According to the above two steps, the given two exponential form equations are simplified as the following two linear equations in two variables.
$(1).\,\,\,$ $2x-y-3 \,=\, 0$
$(2).\,\,\,$ $x+4y-3 \,=\, 0$
According to the first linear equation in two variables, the value of $y$ can be expressed in terms of $x$.
$2x-y-3 \,=\, 0$
$\implies$ $2x-3 \,=\, y$
$\implies$ $y \,=\, 2x-3$
Now, substitute the value of $y$ in the second linear equation in two variables for evaluating the variable $x$.
$x+4y-3 \,=\, 0$
$\implies$ $x+4(2x-3)-3 \,=\, 0$
$\implies$ $x+4 \times 2x-4 \times 3-3 \,=\, 0$
$\implies$ $x+8x-12-3 \,=\, 0$
$\implies$ $9x-15 \,=\, 0$
$\implies$ $9x \,=\, 15$
$\implies$ $x \,=\, \dfrac{15}{9}$
$\implies \require{cancel}$ $x \,=\, \dfrac{\cancel{15}}{\cancel{9}}$
$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, \dfrac{5}{3}$
We have simplified the first linear equation in two variables as $y \,=\, 2x-3$. We can use this equation for solving the variable $y$.
$\implies$ $y \,=\, 2\Bigg(\dfrac{5}{3}\Bigg)-3$
$\implies$ $y \,=\, \dfrac{2 \times 5}{3}-3$
$\implies$ $y \,=\, \dfrac{10}{3}-3$
$\implies$ $y \,=\, \dfrac{10-3 \times 3}{3}$
$\implies$ $y \,=\, \dfrac{10-9}{3}$
$\,\,\,\therefore\,\,\,\,\,\,$ $y \,=\, \dfrac{1}{3}$
Therefore, $x \,=\, \dfrac{5}{3}$ and $y \,=\, \dfrac{1}{3}$ are the solutions of the given exponential equations.
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