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Evaluate $a$ and $b$ if $ {(\sqrt{243})}^{\displaystyle \, a} \div 3^{\displaystyle \, b+1} = 1$ and $27^{\displaystyle \,b} -81^{\displaystyle 4-\frac{a}{2}} = 0$

Two algebraic equations are given and each equation is in terms of $a$ and $b$. The value of $a$ and $b$ are required to find by solving them in this problem.

$ {(\sqrt{243})}^{\displaystyle \, a} \div 3^{\displaystyle \, b+1} = 1$

$27^{\displaystyle \,b} -81^{\displaystyle 4-\frac{a}{2}} = 0$


Simplification of the first equation

Firstly, let us simplify the first given equation.

$ {(\sqrt{243})}^{\displaystyle \, a} \div 3^{\displaystyle \, b+1} \,=\, 1$

$\implies$ $ \dfrac{{\Big(\sqrt{243}\Big)}^{\displaystyle \, a} }{ 3^{\displaystyle \, b+1} } \,=\, 1$

The base of exponential term in denominator is $3$. So, express the $243$ as an exponential term on the basis of number $3$.

$\implies$ $ \dfrac{{\Big(\sqrt{3^5}\Big)}^{\displaystyle \, a} }{ 3^{\displaystyle \, b+1} } \,=\, 1$

The symbol of the square root ($\sqrt{\,\,\,}$) represents the exponent $\dfrac{1}{2}$.

$\implies$ $ \dfrac{{\Big({\Big[3^5\Big]}^\frac{1}{2} \Big)}^{\displaystyle \, a} }{ 3^{\displaystyle \, b+1} } \,=\, 1$

Use power rule of exponent of an exponential term and simplify the numerator.

$\implies$ $ \dfrac{{\Big(3^\frac{5}{2}\Big)}^{\displaystyle \, a} }{ 3^{\displaystyle \, b+1} } \,=\, 1$

$\implies$ $\dfrac{3^\frac{5a}{2}}{3^{\displaystyle \, b+1}} \,=\, 1$

Cross multiply the denominator to shift it to right hand side of the equation.

$\implies$ $3^\dfrac{5a}{2} \,=\, 3^{\displaystyle \, b+1}$

$\implies$ $\dfrac{5a}{2} \,=\, b+1$

$\therefore \,\,\,\,\,\,$ $5a \,=\, 2(b+1)$


Simplification of the second equation

Now, consider the second equation to simplify it.

$27^{\displaystyle \,b} -81^{\displaystyle 4-\frac{a}{2}} = 0$

$\implies$ $27^{\displaystyle \,b} \,=\, 81^{\displaystyle 4-\frac{a}{2}}$

Write the numbers $27$ and $81$ in exponential notation on the basis of number $3$.

$\implies$ ${(3^3)}^{\displaystyle \,b} \,=\, {(3^4)}^{\displaystyle 4-\frac{a}{2}}$

Apply the power rule of power of an exponential term to simplify the equation further.

$\implies$ $3^{\displaystyle \,3b} \,=\, 3^{\displaystyle 4\Big(4-\frac{a}{2}\Big)}$

The bases of the both exponential terms are same. So, the exponents should be equal mathematically.

$\therefore \,\,\,\,\,\,$ $3b \,=\, 4\Big(4-\dfrac{a}{2}\Big)$


Solving both equations

Now, consider the simplified equations and try to solve them to obtain the value of $a$ and $b$.

$5a = 2(b+1)$

$3b \,=\, 4\Big(4-\dfrac{a}{2}\Big)$

There is a $\dfrac{a}{2}$ term in the second equation, the first equation can be expressed as $\dfrac{a}{2}$ to substitute its value in the second equation. So, transform the first equation as follows.

$\implies$ $\dfrac{a}{2} = \dfrac{b+1}{5}$

Now, substitute it in the second equation.

$\implies$ $3b \,=\, 4\Big(4-\dfrac{b+1}{5}\Big)$

$\implies$ $3b \,=\, 4\Big(\dfrac{20-b-1}{5}\Big)$

$\implies$ $3b \,=\, 4\Big(\dfrac{19-b}{5}\Big)$

$\implies$ $5 \times 3b \,=\, 4\Big(19-b\Big)$

$\implies$ $15b \,=\, 76-4b$

$\implies$ $15b+4b \,=\, 76$

$\implies$ $19b \,=\, 76$

$\implies$ $b \,=\, \dfrac{76}{19}$

$\implies$ $\require{cancel} b \,=\, \dfrac{\cancel{76}}{\cancel{19}}$

$\therefore \,\,\,\,\,\,$ $b \,=\, 4$

Now, substitute the value of $b$ in any equation for getting the value of $a$. Here, the first equation is considered.

$5a = 2(4+1)$

$\implies$ $5a = 2 \times 5$

$\implies$ $\require{cancel} \cancel{5}a = 2 \times \cancel{5}$

$\therefore \,\,\,\,\,\,$ $a = 2$

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