Math Doubts

Solve $2^{\displaystyle \large x\normalsize +2} \, 27^{\dfrac{x}{x-1}} \,=\, 9$

There are two exponential functions in the given exponential equation in terms of a variable $x$. It is given that the product of them is equals to $9$. So, the value of $x$ should be evaluated by solving the given exponential equation mathematically.

$2^{\displaystyle \large x\normalsize +2} \, 27^{\dfrac{x}{x-1}} \,=\, 9$

$\implies$ $2^{\displaystyle \large x\normalsize +2} \,\times\, 27^{\dfrac{x}{x-1}} \,=\, 9$

On the left hand side of the equation, the number $2$ raised to the power of $x$ plus $2$ is multiplied by the $27$ raised to the power of the quotient $x$ by $x$ minus $1$. It is given that the product of them is equal to $9$ as displayed on the right hand side of the equation.

Trick to simplify the exponential equation

The base of the second factor on the left hand side of the equation is $27$ and $9$ is the value on the right hand side of the equation. The numbers $9$ and $27$ can be factored in terms of $3$. Hence, it is recommendable to express both quantities in terms of $3$ by the exponentiation.

$\implies$ $2^{\displaystyle \large x\normalsize +2} \,\times\, \Big(3^{\displaystyle 3}\Big)^{\dfrac{x}{x-1}} \,=\, 3^{\displaystyle 2}$

Simplify the equation by math operations

According to the power rule of exponents, the power of an exponential quantity is equal to the base of the exponential quantity raised to the power of product of the exponents.

$\implies$ $2^{\displaystyle \large x\normalsize +2} \,\times\, 3^{\displaystyle \bigg(3 \times \dfrac{x}{x-1}\bigg)}$ $\,=\,$ $3^{\displaystyle 2}$

$\implies$ $2^{\displaystyle \large x\normalsize +2} \,\times\, 3^{\displaystyle \bigg(\dfrac{3 \times x}{x-1}\bigg)}$ $\,=\,$ $3^{\displaystyle 2}$

$\implies$ $2^{\displaystyle \large x\normalsize +2} \,\times\, 3^{\displaystyle \bigg(\dfrac{3x}{x-1}\bigg)}$ $\,=\,$ $3^{\displaystyle 2}$

Observe the quantities on both sides of the equation. The base of the quantity of second factor on the left hand of the equation is same as the base of the quantity on the right-hand side of the equation. Hence, let us try to combine the quantities whose bases are the same and it can be done by shifting the exponential factor on the right side of the equation.

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $\dfrac{3^{\displaystyle 2}}{3^{\displaystyle \bigg(\dfrac{3x}{x-1}\bigg)}}$

According to the quotient rule of exponents, the quotient of exponential quantities is equal to the base of the quantity is raised to the power of difference of the exponents.

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $3^{\displaystyle \bigg(2-\dfrac{3x}{x-1}\bigg)}$

In order to simplify the equation further, find the difference of the fractions by the difference rule of fractions at exponent position on the right side of the equation.

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $3^{\displaystyle \bigg(\dfrac{2 \times (x-1)-3x}{x-1}\bigg)}$

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $3^{\displaystyle \bigg(\dfrac{2 \times x-2 \times 1-3x}{x-1}\bigg)}$

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $3^{\displaystyle \bigg(\dfrac{2x-2-3x}{x-1}\bigg)}$

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $3^{\displaystyle \bigg(\dfrac{2x-3x-2}{x-1}\bigg)}$

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $3^{\displaystyle \bigg(\dfrac{-x-2}{x-1}\bigg)}$

$\implies$ $2^{\displaystyle \large x\normalsize +2}$ $\,=\,$ $3^{\displaystyle \bigg(\dfrac{-(x+2)}{x-1}\bigg)}$

The bases of exponential quantities on both sides of the equation are different. So, it is not possible to solve the value of $x$ by the exponentiation. We can overcome this issue by the logarithmic system. In this problem, we take the natural logarithm on both sides of the equation.

$\implies$ $\log{\Big(2^{\displaystyle \large x\normalsize +2}\Big)}$ $\,=\,$ $\log{\Bigg(3^{\displaystyle \dfrac{-(x+2)}{x-1}}\Bigg)}$

Solve the logarithmic equation by math operations

According to the power rule of logarithms, the logarithm of a quantity in exponential form can be evaluated by multiplying the exponent with logarithm of base of the exponential quantity.

$\implies$ $(x+2) \times \log{2}$ $\,=\,$ $\dfrac{-(x+2)}{x-1} \times \log{3}$

$\implies$ $(x+2) \times \log{2}$ $\,=\,$ $-\dfrac{x+2}{x-1} \times \log{3}$

On both sides of the equation, $x+2$ is a common factor. So, shift the expression on the right side of the equation to the left side.

$\implies$ $(x+2) \times \log{2}$ $+$ $\dfrac{x+2}{x-1} \times \log{3}$ $\,=\,$ $0$

$\implies$ $(x+2) \times \log{2}$ $+$ $\dfrac{(x+2) \times 1}{x-1} \times \log{3}$ $\,=\,$ $0$

$\implies$ $(x+2) \times \log{2}$ $+$ $(x+2) \times \dfrac{1}{x-1} \times \log{3}$ $\,=\,$ $0$

In this equation, $x+2$ is a common factor in both terms of the expression and it can be taken out common from the terms. It transforms the sum of the terms into factor form.

$\implies$ $(x+2) \times \bigg(\log{2}$ $+$ $\dfrac{1}{x-1} \times \log{3}\bigg)$ $\,=\,$ $0$

The simplification process of the equation is completed and it is time to solve the value of $x$ by making each factor equals to zero.

Therefore, $x+2 \,=\, 0$ and $\log{2}+\dfrac{1}{x-1} \times \log{3}$ $\,=\,$ $0$

Let us try to each algebraic equation to solve the value of $x$.

$x+2 \,=\, 0$
$\therefore\,\,\, x \,=\, -2$

It is solved that the value of $x$ is equal to $-2$. Similarly, let’s try to solve the remaining algebraic equation to find another value of $x$.

$\log{2}+\dfrac{1}{x-1} \times \log{3}$ $\,=\,$ $0$

$\implies$ $\dfrac{1}{x-1} \times \log{3}$ $\,=\,$ $-\log{2}$

$\implies$ $\dfrac{1}{x-1}$ $\,=\,$ $-\dfrac{-\log{2}}{\log{3}}$

$\implies$ $\dfrac{\log{3}}{-\log{2}}$ $\,=\,$ $\dfrac{x-1}{1}$

$\implies$ $-\dfrac{\log{3}}{\log{2}}$ $\,=\,$ $x-1$

$\implies$ $x-1$ $\,=\,$ $-\dfrac{\log{3}}{\log{2}}$

$,\,\,\therefore\,\,\,\,\,\,$ $x$ $\,=\,$ $1-\dfrac{\log{3}}{\log{2}}$

$x \,=\, -2$ and $1-\dfrac{\log{3}}{\log{2}}$ are the solutions of the given exponential equation.

Verification

Let us verify each value by substituting it in the given exponential equation.

$2^{\displaystyle \large x\normalsize +2} \, 27^{\dfrac{x}{x-1}}$ $\,=\,$ $9$

Now, substitute $x \,=\, -2$ in the expression on the left-hand side of the equation and evaluate its value.

$=\,\,\,$ $2^{\displaystyle -2+2} \times 27^{\dfrac{-2}{-2-1}}$

$=\,\,\,$ $2^{\displaystyle 0} \times 27^{\dfrac{-2}{-3}}$

According to the zero power rule, the value of $2$ raised to the power zero is equal to one.

$=\,\,\,$ $1 \times 27^{\dfrac{2}{3}}$

$=\,\,\,$ $27^{\dfrac{2}{3}}$

$=\,\,\,$ $27^{\dfrac{1 \times 2}{3}}$

$=\,\,\,$ $27^{\dfrac{1}{3} \displaystyle \times 2}$

$=\,\,\,$ $\Bigg(27^{\dfrac{1}{3}}\Bigg)^{\displaystyle 2}$

$=\,\,\,$ $\Big(\sqrt[\displaystyle \small 3]{27}\Big)^{\displaystyle 2}$

$=\,\,\,$ $\Big(\sqrt[\displaystyle \small 3]{3^{\displaystyle 3}}\Big)^{\displaystyle 2}$

$=\,\,\,$ $3^{\displaystyle 2}$

$=\,\,\,$ $9$

It is evaluated that the value of expression $2^{\displaystyle \large x\normalsize +2} \, 27^{\dfrac{x}{x-1}}$ is equal to $9$ when the value of $x$ is equal to $-2$. Hence, the value $x$ equals to $-2$ is a root of the given exponential equation.

Now, substitute $x$ $\,=\,$ $1-\dfrac{\log{3}}{\log{2}}$ in the expression on the left-hand side of the equation to find its value.

$=\,\,\,$ $2^{\displaystyle 1-\dfrac{\log{3}}{\log{2}}+2}$ $\times$ $27^{\dfrac{1-\dfrac{\log{3}}{\log{2}}}{1-\dfrac{\log{3}}{\log{2}}-1}}$

Simplify the expressions at exponent position in both factors of the expression.

$=\,\,\,$ $2^{\displaystyle 1+2-\dfrac{\log{3}}{\log{2}}}$ $\times$ $27^{\dfrac{1-\dfrac{\log{3}}{\log{2}}}{1-1-\dfrac{\log{3}}{\log{2}}}}$

$=\,\,\,$ $2^{\displaystyle 3-\dfrac{\log{3}}{\log{2}}}$ $\times$ $27^{\dfrac{1-\dfrac{\log{3}}{\log{2}}}{\cancel{1}-\cancel{1}-\dfrac{\log{3}}{\log{2}}}}$

$=\,\,\,$ $2^{\displaystyle 3-\dfrac{\log{3}}{\log{2}}}$ $\times$ $27^{\dfrac{1-\dfrac{\log{3}}{\log{2}}}{-\dfrac{\log{3}}{\log{2}}}}$

Now, split the first factor by the quotient rule of exponents. For adding the fractions in the numerator at exponent position of the second factor, express the number $1$ in terms of $\log{2}$.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{2^{\displaystyle \dfrac{\log{3}}{\log{2}}}}$ $\times$ $27^{\dfrac{\dfrac{\log{2}}{\log{2}}-\dfrac{\log{3}}{\log{2}}}{-\dfrac{\log{3}}{\log{2}}}}$

The quotient of logarithmic quantities can be evaluated in the first factor by using the change of base rule of logarithms.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{2^{\displaystyle \log_{2}{3}}}$ $\times$ $27^{\dfrac{\dfrac{\log{2}-\log{3}}{\log{2}}}{-\dfrac{\log{3}}{\log{2}}}}$

It is time to release the quantity in the denominator of first factor from the logarithm by the fundamental rule of logarithms.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3}$ $\times$ $27^{\dfrac{\dfrac{\log{\Big(\dfrac{2}{3}\Big)}}{\log{2}}}{-\dfrac{\log{3}}{\log{2}}}}$

Now, focus on simplifying the second factor by the mathematical operations.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3}$ $\times$ $27^{\displaystyle -\dfrac{\dfrac{\log{\Big(\dfrac{2}{3}\Big)}}{\log{2}}}{\dfrac{\log{3}}{\log{2}}}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3}$ $\times$ $27^{\displaystyle -\dfrac{\log{\Big(\dfrac{2}{3}\Big)}}{\log{2}} \times \dfrac{\log{2}}{\log{3}}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3}$ $\times$ $27^{\displaystyle -\dfrac{\log{\Big(\dfrac{2}{3}\Big) \times \log{2}}}{\log{2} \times \log{3}}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3}$ $\times$ $27^{\displaystyle -\dfrac{\log{\Big(\dfrac{2}{3}\Big) \times \cancel{\log{2}}}}{\cancel{\log{2}} \times \log{3}}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3}$ $\times$ $27^{\displaystyle -\dfrac{\log{\Big(\dfrac{2}{3}\Big)}}{\log{3}}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3}$ $\times$ $27^{\displaystyle (-1) \times \dfrac{\log{\Big(\dfrac{2}{3}\Big)}}{\log{3}}}$

Now, use the power rule of logarithms for shifting the multiplying factor to exponent position of the quantity in log expression.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times 27^{\dfrac{\log{\Big(\dfrac{2}{3}\Big)^{\displaystyle -1}}}{\log{3}}}$

Use the power rule of logarithms for shifting the multiplying factor to exponent position of the quantity in log expression.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times 27^{\dfrac{\log{\Bigg(\dfrac{1}{\Big(\dfrac{2}{3}\Big)}\Bigg)}}{\log{3}}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times 27^{\dfrac{\log{\Big(\dfrac{3}{2}\Big)}}{\log{3}}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times 27^{\displaystyle \log_{3}{\Big(\dfrac{3}{2}\Big)}}$

Write the base quantity $27$ in exponential form in the second factor.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times \Big(3^{\displaystyle 3}\Big)^{\displaystyle \log_{3}{\Big(\dfrac{3}{2}\Big)}}$

The power of an exponential quantity can be simplified by the power rule of exponents.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times 3^{\displaystyle 3 \times \log_{3}{\Big(\dfrac{3}{2}\Big)}}$

Similarly, the product of a number and a logarithmic quantity can be simplified by the power rule of logarithms.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times 3^{\displaystyle \log_{3}{\Big(\dfrac{3}{2}\Big)^{\displaystyle 3}}}$

Now, use the fundamental rule of logarithms one more time to eliminate logarithm from the second factor of the expression.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times \Big(\dfrac{3}{2}\Big)^{\displaystyle 3}$

Finally, multiply the fractions and evaluate its value.

$=\,\,\,$ $\dfrac{2^{\displaystyle 3}}{3} \times \dfrac{3^{\displaystyle 3}}{2^{\displaystyle 3}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3} \times 3^{\displaystyle 3}}{3 \times 2^{\displaystyle 3}}$

$=\,\,\,$ $\dfrac{2^{\displaystyle 3} \times 3^{\displaystyle 3}}{2^{\displaystyle 3} \times 3}$

$=\,\,\,$ $\dfrac{\cancel{2^{\displaystyle 3}} \times \cancel{3^{\displaystyle 3}}}{\cancel{2^{\displaystyle 3}} \times \cancel{3}}$

$=\,\,\,$ $3^{\displaystyle 2}$

$=\,\,\,$ $9$

It is also calculated that the value of expression $2^{\displaystyle \large x\normalsize +2} \, 27^{\dfrac{x}{x-1}}$ is equal to $9$ when the value of $x$ is equal to $1-\dfrac{\log{3}}{\log{2}}$. Hence, the value $x$ equals to $1-\dfrac{\log{3}}{\log{2}}$ is a zero of the given exponential equation.

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