# Solve $\sqrt{8^0 + \dfrac{2}{3}}$ $=$ ${(0.6)}^{2-3x}$ for the value of $x$

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

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 $x$.

#### Step: 1

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

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

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

#### Step: 2

Write the decimal number $0.6$ in fraction form.

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

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

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

#### Step: 3

$\dfrac{5}{3}$ 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 $\dfrac{1}{2}$.

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

#### Step: 4

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

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

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

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

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

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

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

### Method: 2

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

#### Step: 1

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

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

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

#### Step: 2

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

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

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

#### Step: 3

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

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

#### Step: 4

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

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

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

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

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

#### Step: 5

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.

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

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

#### Step: 6

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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