$\sqrt{8^0 + \dfrac{2}{3}} = {(0.6)}^{2-3x}$

It is an equation which contains exponential terms, a ration number, a decimal number and also a linear equation. They constructed a mathematical relation in the form an equation and it is useful to find the value of $x$ mathematically.

This maths problem can be solved in two different approaches to find the value of $x$.

According to zero exponent rule of the exponentiation, the number $8$ raised to the power of zero is one.

$\implies \sqrt{1 + \dfrac{2}{3}} = {(0.6)}^{2-3x}$

$\implies \sqrt{\dfrac{5}{3}} = {(0.6)}^{2-3x}$

Write the decimal number $0.6$ in fraction form.

$\implies \sqrt{\dfrac{5}{3}} = {\Bigg(\dfrac{6}{10}\Bigg)}^{2-3x}$

$\require{cancel} \implies \sqrt{\dfrac{5}{3}} = {\Bigg(\dfrac{\cancel{6}}{\cancel{10}}\Bigg)}^{2-3x}$

$\implies \sqrt{\dfrac{5}{3}} = {\Bigg(\dfrac{3}{5}\Bigg)}^{2-3x}$

$\dfrac{5}{3}$ is the fraction in left hand side expression but the same term is in reciprocal from in right hand side expression. So, express the $\dfrac{3}{5}$ in its reciprocal from to make terms of both sides have same base.

$\implies \sqrt{\dfrac{5}{3}} = {\Bigg[{\Bigg( \dfrac{5}{3} \Bigg)}^{-1} \Bigg]}^{2-3x}$

Use power law of exponents to simplify the power of an exponential term at right hand side expression.

$\implies \sqrt{\dfrac{5}{3}} = {\Bigg(\dfrac{5}{3}\Bigg)}^{-1 \times (2-3x)}$

$\implies \sqrt{\dfrac{5}{3}} = {\Bigg(\dfrac{5}{3}\Bigg)}^{(-2+3x)}$

The value of square root is an exponent of $\dfrac{1}{2}$.

$\implies {\Bigg(\dfrac{5}{3}\Bigg)}^{\frac{1}{2}} = {\Bigg(\dfrac{5}{3}\Bigg)}^{(-2+3x)}$

The fraction $\dfrac{5}{3}$ is a common base at both sides of the equation. Hence, their exponents should be equal mathematically.

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

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

$\implies 3x = \dfrac{1}{2}+2$

$\implies 3x = \dfrac{5}{2}$

$\implies x = \dfrac{5}{2 \times 3}$

$\therefore \,\,\,\,\,\, x = \dfrac{5}{6}$

This maths question can be solved in another method by squaring both sides to element the square root from left hand side expression.

Take square both sides and eliminate the square root from the left hand side expression.

$\implies {\Bigg[\sqrt{8^0 + \dfrac{2}{3}}\Bigg]}^2 = {\Bigg[{(0.6)}^{2-3x}\Bigg]}^2$

$\implies 8^0 + \dfrac{2}{3} = {\Bigg[{(0.6)}^{2-3x}\Bigg]}^2$

Apply power rule of indices to simply the right hand side expression.

$\implies 8^0 + \dfrac{2}{3} = {(0.6)}^{2 \times (2-3x)}$

$\implies 8^0 + \dfrac{2}{3} = {(0.6)}^{(4-6x)}$

Use zero indices rule to express $8$ raised to the power of zero as one.

$\implies 1 + \dfrac{2}{3} = {(0.6)}^{(4-6x)}$

Express the decimal $0.6$ in fraction from and simply the right hand side expression further.

$\implies \dfrac{5}{3} = {(0.6)}^{(4-6x)}$

$\implies \dfrac{5}{3} = {\Bigg(\dfrac{6}{10}\Bigg)}^{(4-6x)}$

$\implies \dfrac{5}{3} = {\Bigg(\dfrac{3}{5}\Bigg)}^{(4-6x)}$

$\implies \dfrac{5}{3} = {\Bigg(\dfrac{3}{5}\Bigg)}^{(4-6x)}$

In previous method, the fraction $\dfrac{3}{5}$ is expressed as $\dfrac{5}{3}$ by the reciprocal rule of indices but the fraction $\dfrac{5}{3}$ is expressed in reciprocal form in this method.

$\implies \dfrac{1}{\dfrac{3}{5}} = {\Bigg(\dfrac{3}{5}\Bigg)}^{(4-6x)}$

$\implies {\Bigg(\dfrac{3}{5}\Bigg)}^{-1} = {\Bigg(\dfrac{3}{5}\Bigg)}^{(4-6x)}$

The bases of exponential terms at both sides are same. Hence, the indices of them should be equal.

$-1 = 4-6x$

$\implies 6x = 4+1$

$\implies 6x = 5$

$\therefore \,\,\,\,\,\, x = \dfrac{5}{6}$

$x$ is equal to $\dfrac{5}{6}$, is the required solution for this maths problem.

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